[f^ M^^^ Digitiz^d'^^themternet Archive in 2()08 with funding from IVIicrosoft Corporation http://www.archive.org/details/elementarymathemOOyounrich ELEMENTARY MATHEMATICAL ANALYSIS A SERIES OF MATHEMATICAL TEXTS EDITED BY EARLE RAYMOND HEDRICK THE CALCULUS By Ellery Williams Davis and William Charles Brenke. ANALYTIC GEOMETRY AND ALGEBRA By Alexander Ziwet and Louis Allen Hopkins. ELEMENTS OF ANALYTIC GEOMETRY By Alexander Ziwet and Louis Allen Hopkins. PLANE AND SPHERICAL TRIGONOMETRY WITH COMPLETE TABLES By Alfred Monroe Kenyon and Louis Ingold. PLANE AND SPHERICAL TRIGONOMETRY WITH BRIEF TABLES By Alfred Monroe Kenyon and Louis Ingold. ELEMENTARY MATHEMATICAL ANALYSIS By John Wesley Young and Frank Millett Morgan. COLLEGE ALGEBRA By Ernest Brown Skinner. PLANE TRIGONOMETRY FOR SCHOOLS AND COL- LEGES By Alfred Monroe Kenyon and Louis Ingold. THE MACMILLAN TABLES Prepared under the direction of Earle Katmond Hedrick. PLANE GEOMETRY By Walter Burton Ford and Charles Ammerman. PLANE AND SOLID GEOMETRY By Walter Burton Ford and Charles Ammerman. SOLID GEOMETRY By Walter Burton Ford and Charles Ammerman. CONSTRUCTIVE GEOMETRY Prepared under the direction of Earle Raymond Hedrick. JUNIOR HIGH SCHOOL MATHEMATICS By W. L. VoSBURGH and F. W. Gentleman. ELEMENTARY MATHEMATICAL ANALYSIS BY JOHN WESLEY YOUNG PROFESSOR OP MATHEMATICS, DARTMOUTH COLLEGE AND FRANK MILLETT MORGAN ASSISTANT PROFESSOR OF MATHEMATICS, DARTMOUTH COLLEGE THE MACMILLAN COMPANY 1917 All rights reserved Copyright, 1917, By the MACMILLAN COMPANY. Set up and electrotyped. Published July, 1917. Reprinted September, October, 191 7. NoriDoat) }fixtt» J. 8. Gushing Co. — Berwick & Smith Co. Norwood, Mass., U.S.A. PREFACE This book aims to present a course suitable for students in the first year of our colleges, universities, and technical schools. It presupposes on the part of the student only the usual mini- mum entrance requirements in elementary algebra and plane geometry^ The .book has been written with the hope of contributing something toward the solution of the problem of increasing the value and significance of our freshman courses. The recent widespread discussion of this problem has led to the general acceptance on the part of many teachers of certain principles governing the selection and arrangement of mate- rial and the point of view from which it is to be presented. Among such principles, which have guided us in the prepara- tion of this text, are the following. 1. More emphasis should be placed on insight and under- standing of fundamental conceptions and modes of thought, less emphasis on algebraic technique and facility of manipula- tion. The development of proficiency in algebraic manipulation as such we believe has little general educational value. It is valuable only as a means to an end, not as an end in itself. A certain amount of skill in algebraic reduction is, of course, essential to any effective understanding of mathematical pro- cesses, and this minimum of skill the student must secure. But it seems undesirable in the first year to emphasize the formal aspects of mathematics beyond what is necessary for the understanding of mathematical thought. This is espe- cially true for that great majority of students who do not continue their study of mathematics beyond their freshman :i79f40ft vi PREFACE year and who study mathematics for general cultural and dis- ciplinary purposes. It seems to us altogether probable, how- ever, that even in the case of students who expect to use mathematics in their later life work (as scientists, engineers, etc.) greater power will be gained in the same length of time, if their first year in college is devoted primarily to the gain- ing of insight and appreciation, rather than technical facility. Experience has shown only too conclusively that in many cases the emphasis usually placed on formal manipulation effectually prevents the development of any adequate sort of independent power. 2. The reference above to the general cultural and disciplin- ary aims of mathematical study at once raises the question as to the selection of the material that is to form the content of the course. The cultural motive for the study of mathematics is found in the fact that mathematics has played and contin- ues to play in increasing measure an important role in human progress. An educated man or woman should have some con- ception of what mathematics has done and is doing for man- kind and some appreciation of its power and beauty. To this end our introductory courses should cover as broad a range of mathematical concepts and processes as possible. In particu- lar, they must not confine themselves to ancient and medieval mathematics, but must give due consideration to more modern mathematical disciplines. The fundamental conceptions of the calculus must be introduced as early as is feasible in view of the essential role they have played in the progress of civilization. If this broad cultural aim is accepted as one of the funda- mental principles in the selection of material, we shall readily agree that much that is now generally considered necessary can and should be eliminated from our general courses in PREFACE vii mathematics. Almost all of the conventional course in solid geometry would fall in this category, as well as much of what is now taught as college algebra, all of the more specialized portions of analytic geometry, etc. The time thus gained could then be used for topics that are culturally more significant. 3. The disciplinary motive for the study of mathematics is the one most often emphasized and ,need not be elaborated here. In spite of much recent criticism of the doctrine of formal discipline in education and in spite of the fact that some of this criticism as applied to mathematics seems to us justified, we firmly believe that faith in the disciplinary value of mathematics is fundamentally sound. Teachers of mathe- matics need, however, to formulate with precision their aims and purposes in this direction and make their practice conform to this formulation. The disciplinary value of mathematics is to be sought primarily in the domain of thinking, reasoning, reflection, analysis ; not in the field of memory, nor of skill in a highly specialized form of activity. We come back here to the conflict between insight and technique discussed earlier in this preface. Suflice it to remark here that the purpose of technical facility is to economize thought, rather than to stimulate it. If our primary purpose is to stimulate thought, we must place the major emphasis on the mathematical formu- lation of a problem and on the interpretation of the final re- sult, rather than on the formal manipulation which forms the necessary intermediate step ; on the derivation of a formula rather than merely on its formal application ; on the general significance of a concept rather than on its specialized function in a purely mathematical relation. If we desire to enhance the general disciplinary value of mathematics, we will seek out and emphasize especially those conceptions and those modes of thought of our subject which viii PREFACE are most general in their application to the problems of our everyday life. It is fortunate for our purpose — and it is probably more than a mere coincidence — that the conceptions and processes of mathematics which most readily suggest themselves in this connection are the same that are suggested by the cultural motive discussed earlier. The concept of funo- tionality and the mathematical processes developed for the study of functions are precisely the things in mathematics that have most effectively contributed to human progress in more modern times ; and the thinking stimulated by this concept and these processes is fundamentally similar to the thought which we are continually applying to our daily prob- lems. "Functional thinking," to use Klein's famous phrase, is universal. It comes into play when we make the simplest purchase, as well as when we attempt to analyze the most complicated interplay of causes and effects. In the preparation of this text, we have sought to give an introduction to the elementary mathematical functions, the concepts connected therewith, the processes necessary to their study, and their applications. By making the concept of a function fundamental throughout we believe we have gained a measure of unity impossible when the year is split up among several different subjects. The arrangement of this material is exhibited in the table of contents and the text proper, and need not be discussed here. We would merely call attention briefly to some features which seem to require emphasis or explanation. The change in the value of a function due to a change in the value of a variable is emphasized from the very beginning. The change ratio A y/A x is introduced in Chapter III for the linear function, and the derivative is introduced as the slope of the graph of a quadratic function in Chapter IV, although PREFACE ix the word "derivative" is not introduced until Chapter XIX. Derivatives are used in Chapters IV, V, X, XII, XIII, and XIX. We have discussed rather more fully than is customary those topics which involve new and important concepts, and have been correspondingly brief where we felt the student ought to be able to supply the argument himself. We have tried throughout to place the emphasis on an understanding of the general bearing of the principles, and have consistently tried to minimize difficulties of mere algebraic technique. It seems quite likely that customary classroom procedure will, therefore, need to be modified in the direction of lessening the time given to formal recitations and increasing the opportuni- ties for informal discussion. A number of questions have been inserted among the exercises which it is hoped will stimulate such discussion ; this is the purpose also of a num- ber of the " Why's " scattered throughout the text. The lists of "Miscellaneous Exercises" found at the end of chapters beginning with Chapter XI contain some exercises too difficult for assignment in an average class. These may be used to advantage, we hope, in so-called " honor sections " consisting of men who have shown exceptional ability in mathematics. A word regarding our conception as to how the text may be applied to meet the varying mathematical preparation of students will not be out of place. At Dartmouth College we propose to distinguish in this connection only two kinds of freshmen : those who enter without trigonometry, and those who have passed a course in trigonometry in their secondary school. The former will cover the first fifteen chapters of this text in a course meeting three hours per week throughout the year (about ninety assignments). These men will have all the necessary preliminarj^ training for the usual courses X PREFACE in the calculus. Those students who enter with trigonometry will cover the first nineteen chapters in a course meeting three hours per week throughout the year, covering the material of Chapters VI, VII, VIII, and IX (which for them is largely review) in about three weeks. In a course meeting five times per week throughout the year, there should be ample time also for a thorough study of the important topics of Chapter XX (Determinants) and Chapters XXI-XXII (Functions of two independent variables ; analytic geometry of space). So much has been said in recent years in favor of a unified course in mathematics for freshmen that it seems desirable actually to try it out in practice. For this purpose a text- book is necessary. We do not believe that this text will solve the problem ; the most we can hope for is that we have secured a first approximation. It is for this reason that we urgently request users of this text to communicate to us any criticisms or suggestions that occur to them looking to the improvement of possible later editions. In particular, we should like advice and counsel as to the possible desirability of increasing the amount of calculus included in the first year. This could be done by devoting less space to the purely geo- metric aspects of analytic geometry. On theoretical grounds we believe this to be desirable. We felt, however, that we ought to be conservative in case of an innovation of this sort, with a view of seeing how the introduction to this limited extent of the notion of the derivative in the first year fares. If the results are satisfactory, we could then take the next step with confidence. J. W. YOUNG. F. M. MORGAN. Hanover, N. H., AprU, 1917. CONTENTS PART I. INTRODUCTORY CONCEPTIONS Chapter I. Functions and Their Representation Chapter II. Algebraic Principles and Their Connection with Geometry 1-32 Chapter Chapter I. IT. III. Chapter Chapter Chapter Chapter PART II. ELEMENTARY FUNCTIONS III. The Linear Function. The Straight Line IV. The Quadratic Function . Graphs of Quadratic Functions . Applications of Quadratic Functions '. Quadratic Equations .... V. The Cubic Function. The Function jr" VI. The Trigonometric Functions VII. Trigonometric Relations VIII. The Logarithmic and Exponential tions Chapter IX. Numerical Computation . I. Errors in Computation .... II. Logarithmic Solution of Triangles III. The Logarithmic Scale. The Slide Rule IV. Logarithmic Paper .... Chapter X. The Implicit Quadratic Functions . Func 64-97 98-128 98-114 115-119 120-128 129-142 143-187 188-211 212-235 236-264 236-241 242-251 252-259 260-264 PART III. APPLICATIONS TO GEOMETRY Chapter XL The Straight Line 293-319 Chapter XII. The Circle xi xii CONTENTS Chapter XIII. The Conic Sections . 887-876 Chapter XIV. Polar Coordinates . . 877-891 Chapter XV. Parametric Equations 892-401 PART IV. GENERAL ALGEBRAIC METHODS — THE GENERAL POLYNOMIAL FUNCTION Chapter XVI. Miscellaneous Algebraic Methods . . 402-419 Chapter XVII. Permutations, Combinations, and Proba- bility. The Binomial Theorem 420-431 Chapter XVIII. Complex Numbers 482-448 Chapter XIX. The General Polynomial Function. The Theory of Equations .... 449-474 Chapter XX. Determinants 476-498 PART V. FUNCTIONS OF TWO VARIABLES — SOLID ANALYTIC GEOMETRY Chapter XXI. Linear Functions. The Plane and Straight Line 494-618 Chapter XXII. Quadratic Functions. Quadric Surfaces . 614-681 Tables . . . • 684-642 Powers and Roots 534 Important Constants 535 Four-Place Logarithms 536-537 Four-Place Trigonometric Functions .... 538-542 Index 648-648 ELEMENTARY MATHEMATICAL ANALYSIS ELEMENTARY MATHEMATICAL ANALYSIS PART I. INTRODUCTORY CONCEPTIONS CHAPTER I FUNCTIONS AND THEIR REPRESENTATION 1. The General Idea of a Function. Our daily activities continually furnish us with examples of things that are related to one anotherj of quantities which depend on certain other quantities, which change when certain other quantities change. Thus, a man's health is related to the food he eats, the exer- cise he takes, and to many other things. The price of any manufactured article depends on the cost of production, while the latter cost in turn depends on the cost of the raw ma- terial, the cost of labor, etc. The weather depends on a variety of conditions. These are complicated examples of dependence. There are very simple examples. Thus the price paid for a certain quantity of sugar depends on the num- ber of pounds bought and the price per pound ; the area of a square depends on the length of one of its sides ; and so forth. In all such cases, where some quantity depends on some other quantity or quantities, we say that the former is a func- tion of the latter. Thus the price of an article is a function of the cost of production, the area of a square is a function of the length of one of its sides, etc. B 1 2 MATHKMATICAL ANALYSIS [I, § 2 2. General Laws. Many problems of science consist in expressing as accurately as possible one quantity in terms of another quantity on which the first depends. The statements, " The area of a square is equal to the square of the length of one side," and " The speed of a body falling from rest is pro- portional to the time it has fallen " are simple examples. At the basis of this idea of dependence or functionality is the notion of a general law which the quantities in question obey. Most of the problems of civilized life are concerned, directly or indirectly, with the investigation of such laws. Thus medical science seeks to discover the laws governing health, economies' investigates the laws governing the produc- tion and distribution of wealth, the business man studies the conditions which influence his business and his profits. In every case the investigation of the law in question involves finding out how something is related to, depends on, changes with, something else ; i.e. the study of a function of some kind. The ability to think clearly about such relationships is of the highest importance to every one. This course in mathe- matics is primarily concerned with the study of certain of the simpler kinds of functions and their applications. 3. Numbers and Quantities. We shall confine ourselves in general to the study of relations between things which can be ^measured. We can then always speak of the amount of one of them. Such an amount is expressed, in terms of a suitable unit of measure^ by means of a number. Anything that can be represented by means of a number we shall call a quantity. A function expressing the relation of one such quantity to another gives rise to a relation between numbers. A very power- ful aid in studying functions is their geometric representation, which we shall discuss presently. We must consider first, however, the geometric representation of a single quantity. I, § 5] REPRESENTATION OF FUNCTIONS 3 4. The Arithmetic Scale. The distinction between two of the simplest kinds of quantities can be illustrated by reference to their geometric or graphic representation. Every one is familiar with the so-called arithmetic scale (Fig. 1), of which the I i j I \ 1 1 ^ 1 . , . ^ r- J 3 3 4 5 6 Fig. 1 yard stick and tape measure are examples. The divisions of the scale in these cases represent lengths. Another example is the beam on a certain kind of balance ; here the divisions of the scale represent weights. A characteristic feature of an arithmetic scale is that it begins at some point and extends from in one direction. The quantities represented by such a scale are expressed by means of the numbers of arithmetic. These in turn represent simply the magnitude, or the size, or the amount, of something (as 12 yd. of cloth, 96 lb. of sugar, etc.). 5. The Algebraic Scale. Hardly less familiar nowadays is the so-called algebraic scale (Fig. 2). The most familiar ex- n — ' — I — I — I — ' — r^ — I J , I — ' — r -3 -2 •^l/ *1 +2i +3 Fia. 2 ample is probably the scale on an ordinary thermometer. Every one knows the meaning of -f- 10° or — 5°. Such an algebraic scale extends in two opposite directions from some arbitrary point (marked 0) of the scale. The quan- tities represented by the points of such a scale are expressed by means of the so-called real numbers of algebra, such as : ..., - 4, - Vi2, - 3, - i 0, + 1, + ii, -. Such a number represents not merely a magnitude, but rather a magnitude and one of two opposite directions or 4 MATHEMATICAL ANALYSIS [I, § 5 senses. These two opposite " senses " are of various kinds according to tlie quantities considered. They are often ex- pressed by such phrases as : " to the right of " and " to the left of," " above " and " below," " greater than " and " less than," " before " and " after," etc. Thus + 10° of temperature means a temperature 10° greater than the arbitrary temperature which we have agreed to indicate by 0° ; whereas — 5° means a temperature 5° less than the temperature indicated l)y 0°. It should be noted that 0° of temperature does not mean the absence of temperature. 6. Magnitudes and Directed Quantities. We have seen in the last two sections that a number may represent simply a magnitude; or, that a number may represent a magnitude and one of two opposite directions. The numbers of arithmetic serve the former purpose, the positive and negative numbers of algebra serve the latter. Thus the number 5 represents simply a magnitude, such as a distance of five miles between two stations or a period of time of five hours. The numbers -h 5 and — 5 also represent magnitudes of five units ; but they represent more than this. They may tell us, for example, that a station is five miles east of a certain place denoted by and that another station is five miles west of the place denoted by 0, respectively ; or that an event took place five hours after or five hours before a certain event. We may then distinguish two kinds of quantities : (1) mag- nitudes, and (2) so-called directed quaiitities. Examples of the former are : the length of a board, the weight of a barrel of flour, the duration of a period of time, the price of a loaf of bread, etc. Examples of the latter are : the temperature (a certain number of degrees above or below zero), the distance and direction of some point yl on a line from some other point B on the line^ the time at which a certain event , , ,5 , . - """^ , ■ - < . ."^. I, § 7] REPRESENTATION OF FUNCTIONS 5 occurred (a certain number of hours before or after a given instant) ; etc.* Geometrically, the distinction between directed quantities and mere magnitudes corresponds to the fact that, on the one hand, we may think of the line segment AB as drawn from A to B or from B to A, and, on the other hand, we may choose to consider only the length of such a segment, irrespective of its direction. Figure 3 exhibits the geometric representation of 5, 4- 5, and — 5. A segment whose direc- tion is definitely taken account of is called a directed segment. The magnitude of a directed quantity is called its absolute value. Thus the absolute value of — 5 (and also of -h 5) is 5. 7. Further Remarks concerning Scales. Scales, both arith- metic and algebraic, occur in practice in a variety of forms. We have hitherto considered only the simplest form, in M^hich the scale is con- structed on a straight line and in which the subdivisions corresponding to the numbers 1, 2, 3, ... (and in case of the algebraic scale also those corresponding to the numbers — 1, —2, —3, •••) are at equal intervals. Neither of these two conditions is essential. A scale may be constructed on a curved line (a circle, for example, in which case it is sometimes called a dial). Scales are also used in which the intervals between the points representing the whole numbers are not equal. Such a scale is called a non-uniform scale. The scales on some forms of thermometers, on a slide rule (see p. 252), on certain types of ammeters and pressure gauges, etc., may serve as examples of non-uniform scales. The scales discussed in §§ 4, 5 are then to be described more fully as rectilinear and uniform. In the future, unless specifically stated otherwise, a scale will always mean a uniform scale. * We are here considering only magnitudes in one of two opposite directions It is also possible to consider as quantities magnitudes taken in any direction in a plane or in space. Thus a force has a certain magnitude and is exerted in a certain direction ; it could be completely represented by a line segment whose length represents the magnitude of the force and whose direction (shown by arrow- head) represents the direction in which it acts. Such quantities are called vectors. We shall have occasion to refer to them again (Chap. XVIII) . 6 MATHEMATICAL ANALYSIS [I, §8 8. Use of Line Segments to Represent Quantities. Statis- tical Data. A common use of line segments to represent quantities is in connection with the graphic representation of statistical data. The table below, for example, gives the areas of the New England States ; the adjacent figure represents these areas by means of line segments. Area of New England States States Maine Vermont New Hampshire Massa- chusetts Connec- ticut Rhode Island Square Miles 33,040 9,565 9,305 8,315 4,990 1,250 Maine Vermont New Hamp, Massachusetts Connecticut Rhode Island JO 15 Fig. 4 ?o 25 30 Thous.sq.miles The method of constructing such a graphic representation should be clear without further comment. The above areas could also be represented by areas, as in the following figure. Vt. N.H. Mass. E3 Fig. 5 In general, this method of representation is not so serviceable. Why ? I, § 8] REPRESENTATION OF FUNCTIONS 7 EXERCISES 1. From the following table represent graphically by means of line segments the enrolment in Dartmouth College during the years 1901-1916 : '01-'02 '02-'03 '03-'04 '04-'05 '05-'06 '06-'07 "07-'08 '08-'09 686 709 802 857 927 1058 1131 1136 '09-'10 'lO-'ll '11-'12 '12-'13 13-' 14 '14-' 15 '15-'16 1197 1165 1242 1246 1284 1336 1422 Use a convenient unit to represent 100 students (say \ in.). Can you then represent the data with complete accuracy ? Why ? 2. Represent graphically the size of the libraries of the following institutions : No. of Volumes No. of Volumes Harvard 1,180,000 Williams 80,000 Yale 1,000,000 Amherst 110,000 Dartmouth 130,000 Wesleyan 100,000 Brown 115,000 Univ. of Vermont . . 91,000 3. Take the edge of a sheet of paper and mark on it a point A. Place this edge along the segment representing the area of Vt. in Fig. 4, the point A coinciding with the left-hand extremity of the segment. Mark the right-hand extremity by a point B on the paper. Do the same with the segment representing N. H., placing the point B at the left-hand extremity, however, and obtaining a new point C, corresponding to the right-hand extremity. Continue this process for the states Mass., Conn., and R. I. The total segment represents the sum of the areas. Show that Me. has an area almost as great as that of the other N. E. states com- bined. The process just described in the above exercise is known as graphic addition. 4. Describe a similar process for graphic subtraction. 5. Show that the distance between two points of an arithmetic scale can always be found by subtraction. Is the same true for the points of an algebraic scale ? What is the meaning of the sign of the difference ? 6. Two algebraic scales intersect at right angles, the point of intersec- tion being the point of each scale, and the units on both scales being the same. Show how to find the distance from any point on one scale to any point on the other. Would your method still be applicable, if the units on the two scales were different ? Explain your answer. 7. In constructing Fig. 5 what theorem of plane geometry regarding the areas of similar figures is used ? Could the result of Ex. 3 have been readily obtained from the representation in Fig. 5 ? 8 MATHEMATICAL ANALYSIS [I, §9 9. The Investigation of Functions. We are now ready to consider in some detail a few special examples of func- tions, in order to familiarize ourselves with certain gen- eral characteristics a function may possess, with certain methods for the representation and study of functions, and with the terminology. This is desirable before taking up the more systematic study of general types of functions. 10. Example i. Tlie temperature as a function of the time. The temperature at a given place is a function of the time of day. At any given time we can determine the temperature by / 7 'Wednesday I S TJmrsday J 9 Fn 1<j;/ Fig. 6 simply reading an ordinary thermometer. For the meteorolo- gist, however, the actual temperature at any instant is of less importance than the changes in the temperature that take place during a period of time (such as a day, a month, etc.). To trace these changes he must know the temperature at every I, § 10] REPRESENTATION OF FUNCTIONS 9 instant. For this purpose he makes use of a self-recording thermometer. A portion of a record of such a thermometer is given in Fig. 6. The way in which such an instrument works is briefly as follows. The pivoted lever shown in the figure (Fig. 7) carries a pencil point. The mechanism of the instrument causes the pencil end of the lever to rise or fall as the temperature rises or falls, so that if a vertical thermometer scale* were adjusted behind the pencil point we could read off the Fig. 7 temperature on this scale. The pencil point rests against a strip of paper, ruled as in Fig. 6, which is mounted on a drum. Clockwork causes this drum to rotate uniformly at the proper speed. The rulings on the strip of paper now explain themselves. The distance between two successive horizontal lines corresponds to 2° of temperature. The distance between two successive vertical arcs corresponds to two hours. The temperature at any instant can then be read from the record on the strip of paper. The way in which such a record may be used is illustrated by the following questions, which refer to the record of Fig. 6. * Since the pencil moves on an arc of a circle, this vertical scale is con- veniently constructed on such an arc, rather than on a straight line. 10 MATHEMATICAL ANALYSIS [I, § 10 1. What was the temperature at noon on each of the three days given ? 2. What was the temperature at midnight between Wednesday and Thursday ? At 6 p.m. on Friday ? 3. What was the maximum and the minimum temperature on each of the three days, and at what times did it occur ? 4. When was the temperature 50° ? During what periods was it above 50°? 5. How would a stationary temperature be recorded ? A rapidly rising temperature ? A rapidly falling temperature ? 6. By how many degrees did the temperature change on Wednesday from noon to 2 p.m. ? Was this change a rise or a fall ? 7. During what two hours on these three days did the greatest rise in temperature occur ? 8. When did the most rapid rise in temperature occur ? When the most rapid fall ? 9. What was the average rate of increase (in degrees per hour) in the temperature from the minimum on Thursday to the maximum on Thurs- day ? The average rate of decrease from the maximum on Wednesday to the following minimum ? 11. Graphic Representation. In the preceding example we exhibited the temperature as a function of the time by means of a curve drawn with reference to a time scale and a tempera- ture scale. Such a curve is called a graph of the function in question. Such a graphic representation gives a vivid picture of the function ; but it is limited in accuracy. Why ? Can a change in temperature of 0.1° be distinguished on this graph ? 12. Example 2. Speed in terms of the time. Keadings of the speedometer of an automobile taken every five seconds from a standing start are given in the following table : Number of seconds after start 5 10. 15 20 25 30 35 Speed in miles per hour 2 6 7 16 21 28 36 We proceed to construct a graph of the function thus ob- tained, as follows. We take a piece of square-ruled paper and on one of the horizontal lines (which for convenience we draw more heavily) construct a uniform scale to represent the time I, § 12] REPRESENTATION OF FUNCTIONS 11 (Fig. 8). On the vertical lines through the points representing 5, 10, 15, 20, . . . seconds we lay off segments to represent the speeds at the respective instants. This is most conveniently- done by constructing on the vertical line through a scale representing speed in miles per hour. Thus, by reference to the scale indicated in the figure, the point A represents the «y ■ ~ - r ~ ~ ■" ~ ~ ~ — " - ~ " " " ■ ^ / f ^ 1/ / / / </n 4 • W , / /- f f J t 1 -i * . •X ' ^ ' :: J _ 10 no i Seconds after start Speed of an Automobile Fig. 8 corresponding values : 15 seconds and 7 miles per hour. The other points indicated in the figure are now readily located, or " plotted," in similar fashion. The final step in constructing the figure consists in drawing a " smooth curve " through the points. The curve thus obtained may be used as was the tempera- ture curve discussed in the previous example. We might, for example, conclude from this figure that the speed of the car at the end of 23 seconds was probably about 18 ^ miles per hour. 12 MATHEMATICAL ANALYSIS [I, § 12 The necessity of saying " probably ", however, exhibits an essential difference between this example and the former one. In case of the temperature record the temperature at every instant was automatically recorded ; any point of the curve in that example was as significant as any other point. In the present example the only speeds actually measured are those specifically listed in the above table. And yet the conclusion stated above regarding the speed of the car at the end of 23 seconds is justified. Why ? 1. What was the probable speed of the car at the end of 27 seconds? 2. How long did it take the car to pick up from to 30 miles per hour ? 3. The driver probably shifted gears between the 10th and 15th seconds. What can be said of the reliability of the curve during this interval ? 4. How is the steepness of the curve related to the rate at which the speed is increasing ? 6. Is it possible to calculate, by the use of this figure, approximately how far the car traveled during the first 35 seconds ? 13. Variables. It is desirable to introduce at this point a certain terminology. In the preceding examples we have considered temperature and speed as functions of (i.e. de- pendent on) the time. We have considered several different instants of time and the corresponding values of the tem- perature and the speed. Whenever, in a given discussion, we consider a number of different values of a quantity, such as time, or temperature, or distance, or weight, etc., we call such a quantity a variable. In the above examples, the time and the temperature and the speed are all varia- bles ; and, since in the first example we have thought of the temperature as depending on the time, we may speak of the temperature as the dependent variable, of the time as the independe7it variable. It is often more convenient, however, to call the dependent variable simply the function I, § 15] REPRESENTATION OF FUNCTIONS 13 and the independent variable the variable. Thus, in the second example, the speed was the function and the time was the variable. 14. Tabular Representation. Interpolation. In the second example we secured data concerning a function by measurement and exhibited the corresponding values of variable and function by means of a table of values. Such a table is called a tabular representation of the function. The accuracy of such a repre- sentation is limited only by the precision of measurement. Such a table, however, gives an incomplete description of the function. AVhy? The process of obtaining values of the function for values of the variable that lie between the re- corded values stated in the table is called interpolation. When the interpolated values are read from a graph of the function, the process is known as graphic interpolation. The answers to the first two questions at the end of § 12 were obtained by graphic interpolation. 15. Example 3. Volume of ivater as a function of the tem- perature. When 1000 cc. of water at 0° centigrade is heated, it is found that the volume of the water changes according to the following table. Degrees Centigrade Cubic Centimeters 1000.00 2 999.90 4 999.87 6 999.90 8 999.98 Degi-ees Centigrade Cubic Centimeters 10 1000.12 12 1000.32 14 1000.57 16 1000.86 20 1001.61 It requires a rather careful examination of this table to learn that as the temperature (the variable) is increased 'from 0° the volume of the water (the function) decreases and then in- creases. A graphic representation of this function, analogous to the examples already considered in §§ 10, 12, would have yielded this result at a glance. It is our next concern to 14 MATHEMATICAL ANALYSIS [I, § 15 see how such a representation can be constructed, in this case. To this end we a take a piece of square-ruled paper and on one of the horizontal lines construct a uniform scale to repre- sent temperatures. At the points representing 0°, 2°, 4°, 6°, •••, we would then lay off on the vertical lines distances that are 1 1 -^ - - ' - r- r r ■^ f y / 1 JJ / / 1001.0 f / > ' / / / / / / / 1000.0 r ^ /' s «v, ^ f 9 ^-m ^ s 1 999.0 ^ ""* FlO 9 ^ Degrees Centigrade, to represent the volumes in which we are interested. However, at this point a difficulty presents itself. The numbers representing the volumes in question are so large, and the differences between the volumes for the various tempera- tures so small, that, if we choose the unit on the vertical scale small enough to represent these volumes on a sheet of paper of convenient size, it would be a practical impossibility I, § 15] REPRESENTATION OF FUNCTIONS 15 to represent the volumes with sufficient accuracy to make the diiferences in the volumes distinguishable. It is precisely these variations in volume, however, in which we are primarily interested. To overcome this difficulty, we adopt the expedient of ex- hibiting merely that portion of the graphic representation in which we are primarily interested, and are then able to use a largely magnified scale. That is, we observe that all the volumes in which we are interested lie between 999.00 cc. and 1001.00 cc. We may then assume that the points on the line on which we constructed the temperature scale are at a height representing 999.00 cc. (Fig. 9). In other words we suppose the zero point' of the vertical volume scale to be a great dis- tance below the point at which we are working. We construct a portion of the volume scale on the vertical line through 0, marking the latter point 999.0 and choosing the unit on this scale sufficiently large to meet our requirements. In the figure, as drawn, each of the vertical divisions represents 0.1 cc. The construction of the points P, Q, R, ••• is then readily made. A smooth curve drawn through the points thus plotted then gives the graph *of the function. Here, again, the points in the curve between the points given by the table are uncertain ; but the regularity with which the given points are arranged together with the nature of the phenomenon we are considering leaves little room for doubt that, if the volumes for 1°, 3°, 5°, — should be measured and the resulting volumes plotted, the resulting points would be located upon (or at least very near to) the curve drawn. 1; What is the volume of water at 1° ? at 19° ? 2. What is the minimum volume, and at what temperature does it occur ? 3. At what temperature besides 0° is the volume 1000.00 cc? 16 MATHEMATICAL ANALYSIS [I, § 15 EXERCISES 1. The following temperatures were observed at Hanover, N.H., on a certain day in February, 1914 : Midnight -12°F. 1 A.M. -13° 9 a.m. - 12° F. 6 P.M. + 18° F. 2 A.M. -14° 10 A.M. - 2° 6 p.m. + 11° 3 a.m. -15° 11 A.M. + 4° 7 P.M. + 6° 4 A.M. - 17° Noon + 10° 8 p.m. + 2° 5 A.M. -20° 1 P.M. + 12° 9 P.M. + 1° 6 A.M. -21° 2 P.M. + 14° 10 p.m. 0° 7 A.M. -22° 3 p.m. + 19° 11 p.m. - 2° 8 a.m. -19° 4 p.m. + 22° Midnight -' 4° Plot the corresponding points on square-ruled paper, and draw an approximate graph of the function. Assuming this graph to be correct, what was the temperature at 6.30 a.m.? At 6.30 p.m.? What was the total range (the difference between the maximum and the minimum) of temperature during the day ? How long did it take the temperature to rise from its minimum to its maximum ? At what average rate in degrees per hour did the temperature rise during this period ? 2. A stiff wire spring under tension is found experimentally to stretch an amount d under a tension T as follows : r in lb 10 15 20 25 30 d in thousandths of in. . 8 . 12 16.3 '^ 20 23.5 Plot the above data. What would the stretch be when the tension is 12 1b.? 271b.? 23 1b.? 3. The intercollegiate track records are as follows, where d is the dis- tance run and t is the time : d 100 yd. 220 yd. 440 yd. 880 yd. 1 mile 2 miles t' 9^ sec. 2H sec. 48 sec. 1 m. b^ sec. 4 m. 14f sec. 9 m. 23| sec. I, § 15] REPRESENTATION OF FUNCTIONS 17 Plot these records by points in a plane, and draw a smooth curve through them. Are the points of this curve significant ? Why ? What would you expect the record for 600 yd. to be ? For 1600 yd.? For 1000 yd.? Compare the results of these interpolations with the actual records for these distances. 4. The following table shows the distance at which objects at sea- level are visible from certain elevations : Elevation Distance Elevation Distance Elevation Distance Feet Miles Feet Miles Feet Miles 1 1.3 40 8.4 200 18.7 5 3.0 50 9.3 300 22.9 10 4.2 100 13.2 500 29.6 20 5.9 150 16.2 1000 33.4 30 7.2 Plot the graph of this function. Use a different scale for elevation for values from 100 to 1000 ft. from that used from 1 to 50 ft. Why ? 5. The following is an extract of the mortality table prescribed by statute in most states as the basis on which the reserves of life insurance companies shall be computed : Age Number Living Age Number Living Age Number Living 10 15 20 26 * 30 35 100,000 96,285 92,637 89,032 85,441 81,822 40 45 50 55 60 65 78,106 74,173 69,804 64,563 57,917 49,341 70 75 80 85 90 95 38,569 26,237 14,474 5,485 847 3 Draw the mortality curve. Of 100,000 living at the age of 10 years approximately how many would be alive at 32 years ? At 57 years ? How would you represent on the graph the number dying during any given period of five years ? 18 MATHEMATICAL ANALYSIS [I, § 16 16. Empirical Functions and Arbitrary Functions. The examples of functions we have hitherto considered have been taken from observed measurements of relations existing in nature and life about us. Such functions are called empirical. Another type of functions may now engage our attention. They may be called arbitrary or artfjicial The following will serve as an example. 17. Example 4. Letter postage. According to the postal regulations the postage on letters is fixed at two cents per u u 10 s e : 2 3 4 Letter Postage Fig. 10 G Ounces ounce or fraction thereof. The graph showing the relation between the amount of postage and the weight of the letter is then given by Figure 10. 18. Constant Functions. Continuous and Discontinuous Functions. The graph just referred to exhibits two peculiar- ities that we have not yet had occasion to observe in connection with a function. (1) The value of the function may make a sudden jump as the variable passes through certain values (in this case when the weight passes through the values 1 oz., 2 oz., etc.) without taking on the intermediate values. In the present case, as the weight is increased from exactly 1 oz. to the slightest amount I, § 19] REPRESENTATION OF FUNCTIONS 19 above 1 oz. the postage jumps from 2 cents to 4 cents. A function with such breaks, or changes of a definite amount for no matter how slight a change in the variable, is said to be discontinuous for those values of the variable at which the break or jump occurs. A function, on the other hand, whose graph is a continuous line or curve without such sudden breaks or changes is said to be a continuous function.* (2) Portions of this graph are horizontal straight lines, which means that certain changes in the variable produce no corre- sponding change in the value of the function. Thus, the postage does not change as the weight of the letter is in- creased from slightly more than 1 oz. to 2 oz. In such a case we say that the function is constant (or stationary) for the interval of the variable in question. We should observe, further, that the graph of the function as drawn does not furnish a unique value for the function at the points of discontinuity, i.e. when the weight is 1, 2, 3, ••• oz., since there is nothing to indicate whether we should take the lower or the higher value. As a matter of fact the arbi- trary definition of the function specifies that the lower value is to be taken. 19. More about Arbitrary Functions. We must not assume, of course, from the preceding example that every arbitrary function is discontinuous. In fact, we should note that if we take any square-ruled paper, construct on it a horizontal scale, any number of which we will designate by x, and a vertical scale, any number of * The word continuous is used in mathematics in a highly technical sense, the full discussion of which is beyond the scope of an elementary course. The definition of the term given above is sufficiently precise for our present purposes. Later we shall have more to say of it. 20 MATHEMATICAL ANALYSIS [I, § 19 which we will call y, and then draw an arbitrary curve across the paper, as in Fig. 11, we thereby define a relation between the numbers x of the horizontal scale and the numbers y of the vertical scale, such that to every value of x corresponds a certain value (or possibly a set of values) of y ; i.e. we define y as a function of x.* The .reason for the phrase in paren- 5eii - ... -- ■ w - " " ' t '-'- - :: '^ : ft :- ±:i L r . ! ■, iij • 1 is , . " "'4'" . . . . . . • . ::: i:: -^\'.:::i : 5 T- i^^ij >4 --■' : ir T^l ii M - ±i ; Fig. 11 Fig. 12 theses in the last sentence is as follows. If the curve we draw is such that for any value of x the corresponding vertical line cuts the curve in more than one point, there will be associated with such a value of x more than one value of y (Fig. 12). The variable y is in such a case still a function of X, since the values of y are determined by the values of x. The distinction between functions of the latter type and those previously considered is made by the following definitions : If to every value of the variable under consideration there corresponds a single value of the function, the function is said to be single-valued or one-valued. If to any value of the vari- able corresponds more than one value of the function, the latter is said to be multiple-valued. *The accuracy with which a function is defined by its grapb depends on th6 accuracy with which it is possible to read the two scales of reference and the ** fineness " of the curve. I, § 20] REPRESENTATION OF FUNCTIONS 21 We shall for the present be concerned primarily with one- valued functions only, although one example of a two-valued function will occur soon. Multiple-valued functions will be considered later (Chapter X). EXERCISES 1. From the following data construct a graph showing the cost of domestic money orders in the United States : Amount of Order Kate Amount of Okder Katk, • Not over $ 2.50 3 cents Over $30.00 to | 40.00 15 cents Over .| 2.50 to |5.00 5 cents Over 40.00 to 50 00 18 cent^ Over 5.00 to 10.00 8 cents Over 50.00 to 60.00 20 cents Over 10.00 to 20.00 10 cents Over 60.00 to 75.00 25 cents Over 20.00 to 30.00 12 cents Over 75.00 to 100.00 30 cents 2. Draw a figure showing the rates for parcel-post packages for zone 1 ; for zone 2 ; for zone 3. Compare these graphs. 3. Draw a figure to represent the cost of gas in your own city. Is there a different rate for large consumers ? If so, will this show clearly on the graph ? How ? 4. On a piece of square-ruled paper draw graphs of continuous func- tions which are rapidly increasing ; rapidly decreasing ; slowly increas- ing ; slowly decreasing. 5. Draw the graph of an arbitrary function which is increasing and in which the rate at which it increases is increasing. Also that of an in- creasing function in which the rate of increase is decreasing. 20. Analytic Representation of Functions. We have hitherto considered two methods of representing a function, the graphic and the tabular. There is a third method, called the analytic, which in its simplest form consists of the expression of the function in terms of the variable by means of d, formula, from which the corresponding values of the variable and the function can be computed. The following will serve as examples. 22 MATHEMATICAL ANALYSIS [I, § 21 21. Example 5. Capital and interest. The amount A in t years of $ 1000 drawing simple interest of 5 % is given by the formula (1) ^ = 1000 + 50<. By substituting for t a suc- cession of values and com- puting the corresponding values of A^ we obtain from this formula a tabular rep- resentation of the function. This in turn can be repre- sented graphically. The a 1300 1250 1200 1150 1100 1050 1000 p n -r-i n - -1 - - - - ■ - - " ■ " ~ - ' " ' - - - ■■ - - ■ ^ ' =: ' ^1 » ■ - , •f ■ -' / < . i V 6 Y e n ra t (years) A (dollars) 1000 1050 1100 1150 1200 1250 1300 table above and Fig. 13 are the result.* The points plotted ap- pear to be on a straight line. Prove that they are. 22. Example 6. TJie area of a square. The area (in square inches) of a square whose side is x inches long is given by the formula y = x2. From this equation, we readily compute the following table. ^ Jl m -t -. 16 /- t 7 19 ~J- 12 2 "F 5? y _ _ _ ^ s t - - ~ y A_ , Z ^ 4- 7 .<^ 0^ 1 2 s 4 '0 Inches Fig. 14 X (in.) . . 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 y (sq. in.). 0.25 1.00 2.25 4.00 6.25 9.00 12.25 10.00 * In practice bankers do not take account of fractions of a day in comput- ing interest. Strictly speaking, therefore, the graph of the function A, as used in practice, is discontinuous. This practice of bankers is, however, dic- tated by convenience. It does not alter the fact that the function, as such, is continuous. 1, § 24] REPRESENTATION OF FUNCTIONS 23 Using these values, it is now easy to draw the graph, which is shown in Fig. 14.* X M. *iL Fig. 15 23. Example 7. The function de- fined by a circle. It is often desirable to obtain an analytic representation of a function, originally given graphically or by means of a table. Such an analytic representation is sometimes easy to obtain. Suppose, for example, that on square-ruled paper an a:-scale and a ?/-scale have been constructed with the units on the two scales equals and suppose that with the common 0-point of the scales as a center a circle is drawn with a radius of 2 units (Fig. 15). The functional rela- tion between the variables x and y defined by this curve is to be expressed by means of a formula. If P is any point on the circle, the absolute values of the x and the y of this point form the legs "of a right-angled triangle of which the hypotenuse measures 2 units. By a well-known theorem of geometry we have then y2 = 4 — a;2 or y =± V4 — x^. This is the analytic representation sought. It may be noted that we have here to do with a two-valued function. 24. Range of a Variable. We had occasion some time ago (§ 13) to introduce the term variable. In the future such a quantity will generally be represented by a symbol, such as a;, or 2/, or t, etc. Indeed this was done in some of the preceding examples. The various values attached to such a symbol throughout the discussion are numbers. These numbers con- stitute the range of the variable in question. The range of a variable is usually determined by the nature of the problem under consideration. Often it is very definitely restricted. Thus in the case discussed in the last article the * When, as here, the only fractional parts of a unit which occur are halves, quarters, etc., it is convenient to use a ruled paper on which the larger units are subdivided into four or eight parts instead of ten. 24 MATHEMATICAL ANALYSIS [I, § 24 range of x (as well as that of y) consists of all (real) numbers from — 2 to + 2, and no others. For numbers outside this range, the function in question is not defined. Again, in the case of the mortality table considered in Ex. 5, p. 17, the range of the dependent variable (the number of persons living at a given age) is restricted to whole numbers less than 100,000 ; fractional values of the variable are here meaningless. 25. Increasing and Decreasing Functions. A function which increases when the variable increases is called an in- creasing function ; if, on the other hand, the function decreases as the variable increases, the function is called decreasing. Thus the amount A of capital and interest recently considered is an increasing function of the time t, throughout the range of the latter. Also, the area of a square is an increasing function of the length of one of its sides. On the other hand, the num- ber of people livhig at a given age is a decreasing function of the age. A function may be increasing for certain values of the variable and decreasing for certain other values. Thus, the temperature is during certain portions of the day an increasing function, during other portions a decreasing func- tion. The volume considered in § 15 is a decreasing function of the temperature T, from T = to T = 4, and an increasing function for values of T greater than 4.* If the two scales with reference to which the grapli of a function is constructed are jjlaced in the more usual way, so that the numbers on the scales increase to the right and upward, respectively, what distinguishes the graph of an increasing function from that of a decreasing one ? * In the case of the circle discussed in § 23, the function has two " branches " in the interval from x = — 2 tox=4-2, the one consisting of the positive values of y, the other of the negative values of y. The function may be con- sidered as consisting of two one-valued functions, one of which increases from a;= — 2 to a; = and decreases from x = to x=-\-2, while the other de- creases from x=— 2 to a; = and increases from a; = to x=-\-2. 1, § 25] REPRESENTATION OF FUNCTIONS 25 EXERCISES 1. If a body falls from rest, its speed v in feet per second at the end of t seconds is given by the relation v = 32 t. Construct the graph of v as a function of t. 2. The charge for printing n hundred circulars of a certain kind is |) = 2 n + 10 dollars. Represent the function graphically. 3. The express rate r on a package is computed from the following formula : r =-^(p — 30)+ 30, where w is the weight of the package in 100 pounds and^ is the charge per hundred pounds. Draw the graph of r as a function of w, for each of the values p = 40, 60, 80, 100. What com- ment would you make on this rule for p = 30, or for values of p less than 30 ? This is an example in which the range of the variable is arbitrarily limited to be not less than a certain amount. The formula in this exer- cise really gives r as a function of the two variables w and p. 4. When a body is dropped from a height of 200 ft., its distance s from the ground at the end of t sec. is given by s = 200 — 16.1 1^. Draw the graph of s as a function of t. In how many seconds will the body reach the ground ? At what time is the speed of the body greatest ? Least ? What relation has the steepness of the graph to the speed of the body ? Why ? What are tlie natural limitations on the range of the variable ? 5. In Fig. 13, the beginning of the J.-scale does not appear on the graph. Why ? 6. Rate of increase. In the function of § 21, when < = 2, we have A = 1100. Starling with this initial value of t, let f be increased by 1, by 2, by 3, ••• The corresponding values of A (i.e. the values of A when ^=2 + 1 = 3, 2 + 2=4, etc.) are respectively 1150, 1200, 1250, •■., and the corresponding increases in A over the initial value 1100, are 50, 100, 150, •••. We see then that for these values the increase in ^ is always equal to 50 times the corresponding increase in t.* Show that the same is true if we start with a different initial value of f, say t = 3. Prove, in general, that starting with any particular value, say t '= ti, of t, and any increase in t, say an increase equal to h, that the resulting increase in A is equal to 60h; i.e. that the ratio increase in A ^q corresponding increase in t * When a change in the value of the variable produces a certain change in the value of the function, these two changes correspond to each other. We may then speak of either change as corresponding to the other. 26 MATHEMATICAL ANALYSIS [I, § 25 7. From the result of Ex. 6, show that the graph of the function there considered is a straight line. 8. Make an investigation similar to that in Ex. 6 for the function y=x^ considered in § 22 ; i.e. calculate the increase in y due to an increase in X, under a variety of conditions. For example, let x = 2 initially, and calculate the increases in y resulting from increases of 0.5, 1.0, 1.5, 2.0 in X. For each case calculate the ratio : increase in y corresponding increase in x Is this ratio constant ? Is the increase in y due to an increase in x of 1.0 the same when the initial value of x is 3 as it is when the initial value of x is 2 ? How is the change in the steepness of the graph related to your result ? 9. A car begins to move and gradually increases its speed in such a way that in x sec. it has traveled y = x^ ft. Interpret in this new setting the "increase in y due to a certain increase in x," as computed in the preceding exercise. Show in particular that the "increase in y" is the distance traveled by the car during the interval of time represented by the corresponding " increase in x," and that the ratio increase in y corresponding increase in x is the average speed of the car during this interval. Does this suggest a method for computing approximately the speed of the car at a given instant ? 10. A certain function y has the value 0, when the variable x is 0, and has the value 4, when x = 2. The graph of the function is a straight line. Draw the graph and tabulate, from the graph, the values of y when x=l, 3, 4, 5, C. What is the algebraic relation between y and x ? 11. The graph of a certain function is a straight line. Draw this graph, knowing that y = 0, when x = — 1, and that y = 4, when x = 3. Discover the equation connecting y and x. 26. Statistical Graphs. One of the most generally familiar uses of the graph is in connection with the representation of statistical data. The figure below represents the enrolment in Dartmouth College during the years 1905-1915. The method of its construction should be clear without further ex- planation. I, § 26] RPJPRESENTATION OF FUNCTIONS 27 An essential difference between this sort of graph and those previously considered must, however, be noted. Strictly speak- ing, the graph consists only of the points forming the corners -\ U-4-^-f--r-[-f| > H--J -^ H 1 ^ 1400 --T - " -"f- T :--: z-'-'--/ :::::::::::::::::: ,,-::„:^^: :::::::: „ 1300 J -r ± ±'-:^rk^ - ^ ::::::::::::::::::::::::::::ii-"::L4::::::::: :::::::::: S 1200 ;H; -p^ 1 - - CO ^ r**''' "^ 1 1 1 1 1 1 1 1 li^44Rfem 1 1 1 1 1 1 ! 1 1 1 mm 1 1 1 1 1 1 1 1 1 nTrrl ^ 1100 2 "'" X " 1 ::::>^::::;^:^:._± ^ ^"^l \y(i[[\\\\\ III 900 1905 1906 1907 1908 1909 1910 1911 1912 1913 19li 1915 Years Enrolment of Dartmouth College, 1905-1915 Fig. 16 of the broken line in the figure. The dates, 1905, 1906, ••• refer to the beginning of the college year in September of the years given, and the points plotted give the enrolment at the begin- ning of each such year. The straight lines joining these points are drawn merely for convenience, as an aid to the eye in follow- ing the changes in the enrolment from year to year. The points of these lines between the end points have no significance. The range of the variable here consists of the finite number of dates, 1905, 1906, •••, 1915; and the function considered is discontin- uous. In such a graph interpolation is obviously impossible. Questions (1) During what periods did the enrolment increase ? decrease ? (2) What was the percentage of increase during the 11 years ? (3) What was the average rate of increase (in students per year) from 1905 to 1915? 28 MATHEMATICAL ANALYSIS [I, § 26 (4) If the first point of the graph (1905) be joined to the last point (1915) by a straight line (see figure), how is the steepness of this line re- lated to the average rate of increase ? EXERCISES 1. The maximum temperatures (in degrees Fahrenheit) at Hanover, N.H., on successive days from Oct. 1 to Oct. 15, 1914, were respectively as follows : 59.6, 74.8, 79.7, 82.1, 78.9, 66.6, 61.4, 73.7, 82.5, 73.2, 78.9, 66.8, 55.0, 67.0, 63.5. Construct a graph representing these data by a broken line. Is inter- polation possible ? Why ? 2. American shipping statistics give the total iron and steel tonnage built in the U.S. for the years 1900-1914 as follows : Year Tonnage Year Tonnage Year Tonnage 1900 196,851 1905 182,640 1910 250,624 1901 202,699 1906 297,370 1911 201,973 1902 280,362 1907 348,555 1912 135,881 1903 258,219 1908 450,017 1913 201,665 1904 241,080 1909 136,923 1914 202,549 Draw the graph. Is interpolation possible ? Why ? 27. Summary. As has already been sufficiently indicated, the object of our work thus far has been to make clear the con- cept of a function. To this end we have considered a variety of special functions. Confining ourselves at present to the con- ception of what we have had occasion to define. as a single- valued function of one variable, we have seen that the essential characteristic of such a function may be defined as follows : A variable y is said to be a function of another variable x, if when a value of x is given, the value of y is determined. A variable is a quantity which throughout a given discussion assumes a number of different values. The values which a I, § 28] REPRESENTATION OF FUNCTIONS 29 variable may assume constitute the range of the variable in question. The range of a variable may be limited or not according to circumstances. We have become acquainted with three methods of repre- senting a function : the analytic, the tabular, and the graphic. We have made a beginning in the classification of functions : single-valued and multiple-valued functions ; continuous and dis- continuous functions ; increasing and decreasing functions ; functions of one variable and of more than one variable. We have had occasion to note some of the questions that may arise in the consideration of a function : To determine the value of the function when the value of the variable is given ; the converse problem, to determine the value (or values) of the variable, corresponding to a given value of the function. Both of these problems may involve the process of interpolation. The maximum or minimum value of a function (and the value of the variable for which this maximum or minimum occurs) is often of importance. So also is the rate at which a function changes its values. This, we have seen, is in- timately connected with the steepness of the graph of the function. 28. Algebra as a Tool. The methods to be used in the future for the study of functions and their applications group themselves naturally under three headings corresponding to the methods of representing a function : graphs, analysis, tables. The first of these we have already considered. It has the advantage of presenting the variation of the function vividly to the eye ; in this respect it is the superior of either the tabular or the analytic method of representation. It lacks 30 MATHEMATICAL ANALYSIS [I, § 28 precision, however, since any graph drawn on a piece of paper is in the nature of the case an approximation.* The analytic representation by means of a formula we have touched only very briefly. One of its chief advantages is that of the utmost precision and conciseness. This very conciseness, however, tends to obscure the properties of the function. The tools which enable a sufficiently skillful operator to bring out the hidden properties inherent in a formula are comprised in what is known as mathematical analysis, of which the processes of elementary algebra form the foundation. The more important functions have been tabulated. Such tables are used primarily to facilitate numerical computations. We shall have occasion to use tables frequently. The next chapter is devoted to a brief discussion of certain algebraic processes and of their relation to the graphic rep- resentation already discussed. QUESTIONS FOR REVIEW AND DISCUSSION 1. Give examples from your own experience of quantities that are functionally related. In each case, state as many properties of the function as you can (continuous or discontinuous, increasing or decreasing, etc.). 2. State some general laws and discuss the functional relations they illustrate. 3. Would it be desirable to define a function as follows : y is a function of x, if y changes its value whenever the value of x changes ? Why ? 4. Give, from your experience, concrete examples of the use of an arithmetic scale. Of an algebraic scale. What are the distinguishing characteristics of these two scales ? 5. Describe the three methods of representing a function and discuss the advantages and disadvantages of each. 6. If the graph of a function y of x is a straight line, and the value of the function is known for a; = 4 and for x = 5 (say these values are 20 and 26, respectively), how can the value of the function for aj = 4.5 be calculated (not read from the graph) ? For x = 4.2 ? For a: = 6.7 ? *0n the other hand, we can conceive, theoretically, of a graph which is en- tirely accurate. I, § 28] REPRESENTATION OF FUNCTIONS 31 MISCELLANEOUS EXERCISES 1. The following table gives the pressure of wind in pounds per square feet in terms of the velocity of the wind in miles per hour : Miles per hour 5 10 15 20 30 40 50 60 70 80 Lb. per sq. ft. 0.1 0.5 1.1 2.0 4.4 7.9 12.3 17.7 24.1 31.5 Represent the function graphically. Determine approximately the velocity which will produce a pressure of 10 lb. per square feet. What does the increasing steepness of the curve signify ? 2. The following table, prepared by the U.S. "Weather Bureau, gives the average monthly values of relative humidity at the stations given : C3 < 5 >< •-5 6 -< 5 o > o p New York 75 74 71 68 72 72 74 75 76 74 75 74 Chicago . . 82 81 77 72 71 73 70 71 70 72 77 80 New Orleans . 79 80 77 75 73 77 78 79 77 74 79 79 San Francisco 80 78 78 78 79 80 84 86 81 79 77 80 Plot on the same sheet of paper. Is interpolation possible ? Why ? 3. The following table gives the average weight of men and women for various heights : Height . Weight in Lb. Height Weight in Lb. Men Women Men Women 5 ft. 5 ft. 2 in. 5 ft. 4 in. 5 ft. 6 in. 128 131 138 145 115 125 135 143 5 ft. 8 in. 5 ft. 10 in. 6 ft. 6 ft. 2 in. 154 164 175 188 148 160 170 Represent the two sets of data on the same paper and draw any conclu- sions that seem reasonable. Is interpolation possible ? Why ? 4. The attendance at a base ball park on successive days was as follows : 1002, 1800, 1875, 1375, 1500, 2750, 3520. Represent these data by points in a plane. Is a curve drawn through these points of any significance ? Explain your answer. 32 MATHEMATICAL ANALYSIS [I, § 28 5. The London Economist gives the following table showing the net tonnage of steamships and sailing vessels on the register of Great Britain and Ireland from 1840 to 1912 : Year Steamshii' Sailing Vessel Year Steamship Sailing Vessel 1840 1860 1880 1900 87,930 464,330 2,723,470 7,207,610 2,680,330 4,204,360 3,851,040 2,096,490 1909 1910 1911 1912 10,284,810 10,442,719 10,717,511 10,992,073 1,301,060 1,112,944 , 980,997 902,718 Represent these data graphically on the same sheet of paper. What fact does this graph vividly portray ? 6. The temperature drop t below 212° at which water will boil at differ- ent elevations and the elevation h in feet above sea level are connected by the relation h =1"^ + 517 1. Construct a table of values of h ior t = 0, 5, 10, 15, 20, 25, 30, and draw the graph of 7i as a function of t. At what temperature will water boil on Pike's Peak, 14,000 feet above sea level ? About how high is it necessary to go in order that water will boil at 200"^ ? CHAPTER II ALGEBRAIC PRINCIPLES AND THEIR CONNECTION WITH GEOMETRY 29. Numbers and Measurement. We have already had occasion to distinguish between two kinds of numbers : (a) Numbers each of which represents a magnitude only ; (6) Numbers each of which represents a magnitude and one of two opposite senses, i.e. the so-called signed numbers. It seems desirable at this point to recall the familiar classifi- cation of these numbers and the way in which they serve to give the measures of magnitudes. We confine ourselves first to the numbers of Type (a) above. Integers. The first numbers used were the so-called whole numbers or integers, 1, 2, 3, 4, ..., which represent the results of counting and answer the ques- tion : How many ? They also represent the results of measure- ments, when the magnitudes measured are exact multiples of the unit. The Rational Numbers. When the magnitude measured is not an exact multiple of the unit of measure, other num- bers called fractions must be used. •' A\ 1 1 1 1 *B These numbers are intimately asso- c^ . 1 1 ,/> ciated with the idea of a ratio. ^' ^^ Fig. 17 Thus, in geometry, two line seg- ments AB and CD are called commensurable, if there exists a third segment PQ of which each of the other two is an D 33 34 MATHEMATICAL ANALYSIS [II, § 29 exact multiple (Fig. 17). PQ is then called a common measure of AB and CD. If AB is exactly m times PQ and CZ> is exactly n times PQ, m and n being integers, we say that the ratio of AB to CD is m/n, and we write AB^m CD n' If CD is the unit of length, we have the measure ofAB — — • A number which can be written as a fraction in which the numerator and denominator are both integers is called a rational number* Such numbers suffice to represent the measure of any magni- tude which is commensurable with the unit of measure. The Irrational Numbers. If two magnitudes have no common measure, they are called incommensurable. Thus we know from our study of geometry that the diagonal of a square (Fig. 18) is not commensurable with one of its sides.f Hence, the length of the diagonal of a square whose side is 1 unit cannot be expressed exactly by any rational num- ber. To meet this deficiency the so-called irrational numbers, such as the V2, were introduced. It is beyond the scope of this book to treat irrational num- bers fully. But we may note that they serve to express the * Observe that according to this definition the rational numbers include the integers. The number "zero" is also classed among the rational num- bers. See § 30. t If AB and AC had a common measure /, such that AB = m x I and AC = nX I, where ?n and n are integers, it would follow that n^ = 2in'^ ; but this relation cannot hold for any integers in and «. Why ? II, § 29] ALGEBRAIC PRINCIPLES 35 ratio of pairs of incommensurable magnitudes, and, in particular, to express the measure of any magnitude which is incommensur- able with the unit. Moreover, any irrational number may be rejjresented approxi- mately by a rational number with a7i error which is as small as we please. This follows from the following con- siderations. It is important to note that the result of any actual direct measurement is always a rational number. For example, in measuring a distance, we use a foot rule marked into fourths, or eighths, or thirty-seconds of" an inch, or else some more accurate instrument divided into hundredths or thousandths of a unit, and we always observe how many of these divisions are contained in the length to be measured. The result is, therefore, always a rational number m/n where n represents the number of parts into which the unit was divided. Any such actual meas- urement is, of course, an approximation. The greater the ac- curacy of the measurement (and this accuracy depends among other things on the number of divisions of the unit) the closer is the approximation. Since we may think of the unit as divided into as many divisions as we please, we may conclude that any magnitude can be expressed by a rational number to as high a de- gree of accuracy as may be desired. Thus, the length of the diagonal of a square whose side measures 1 in. is expressed approximately (in inches) by the following rational numbers : 1.4, 1.41, 1.414, 1.4142. These decimals are all rational ap- proximations, increasing in accuracy as the number of decimal places increases, to the irrational number V2.* * Surds, i.e. indicated roots of rational numbers, are not the only irrational numbers. The familiar n = 3.14159 ••• is an example of an irrational number which is not expressible by means of any combinatiou of radicals affecting rational numbers. 36 MATHEMATICAL ANALYSIS [II, § 30 30. The Number System of Arithmetic. The (unsigned) rational and irrational numbers, together with the number zero (which is counted among the rational numbers), constitute the number system of arithmetic. 31. The Nimiber System of Algebra. Corresponding to any unsigned number a (except 0) there exist two signed numbers + a and — a. The magnitude represented by a signed number is called the absolute value of the number, and is indicated by placing a vertical line on each side of the number. Thus the absolute value of + 5 and of — 5 is 5 ; in symbols, |+5|=| — 5|=5. The signed numbers are called rational or irrational accord- ing as their absolute values are rational or irrational. The entire system of positive and negative, rational and irrational, numbers and zero * is called the 7'eal number system and any number of this system is called a real number. These numbers are contained in the so-called number system of algebra.^ * Note that zero is neither positive nor negative. It has no sign. t The number system of algebra contains also the so-called imaginary or complex numbers, which will be discussed later. It may be noted that the words rational, irrational, real, imaginary, are here used in a technical sense. The popular meanings of the terms have no significance. V2 is no more " irrational " (i.e. absurd or crazy) than the number 2 ; and the im- aginary numbers are just as "real" in the popular use of the term as are the (technically) real numbers. Historically, the reason for the use of these words is, however, connected with their customary meaning. For, while the integers and rational numbers are of great antiquity, the irrational numbers were not hitroduced until about the fifteenth century a.d., although incom- mensurable ratios were discussed by the ancient Greeks. At that time their nature was not thoroughly understood, and it was not unnatural then to designate them as irrational. Similar remarks could be made about the introduction of the imaginary numbers toward the end of the eighteenth century. We may add that what we now call "negative" numbers were in the fifteenth century often referred to as " fictitious numbers." II, § 32] ALGEBRAIC PRINCIPLES 37 32. Geometric Representation. Coordinates on a Line. It follows from § 29 that the rational and irrational numbers are just sufficient to express the length of any line segment. Every segment on a line having one extremity at a given point or origin can be represented by such a number; and every such number will determine a definite one of these segments, the unit of measure having been previously chosen. This leads at once to the idea of an arithmetic scale, if we confine ourselves to the numbers of arithmetic, and to the idea of an algebraic scale, if we choose one of the directions on the line to be positive, and use the real numbers of algebra to represent the (now) directed segments. In the future we shall generally confine our discussion to the algebraic case. No confusion need arise from regarding an arithmetic scale as the positive half of an algebraic scale, nor from regarding the numbers of arithmetic as equivalent to the positive numbers (and zero) of the real number system.* It is often convenient to regard the number x which origi- nally represented the length and the direction from to a P Fig. li) point P of the line as representing the point P itself, in which case we call x the coordinate of P (Fig. 19). When we have chosen a point as origin, selected a unit of length, and specified which of the two directions on the line is positive, we say that we have established a system of coordinates on the line. When this has been done, every point P of the line is represented by a number, and every real number represents a definite point of the line. * For this reason we shall often omit the + sign in writing a positive number ; e.g. write simply 5 for + o. The context will always tell whether the number in question is signed or not. 38 MATHEMATICAL ANALYSIS [II, § 33 33. Coordinates in a Plane. We may now give the precise mathematical formulation of the process already used (in con- nection with the construction of the graphs of functions) for " plotting " points in a plane. The essential features of this process are as follows (Fig. 20). We locate arbitrarily in the ^ Secona quadrant First quadrant + 5 ,Pi ^ M.O t/, -^ 31, X' -5 ; " , J X quadrant ^s .5 ■ Pi Third Fourth quadrant Y' Fig. 20 plane two algebraic scales, a horizontal one called the x-axiSj and a vertical one called the y-axis. These two scales, called the axes of reference^ intersect in the zero point of each scale ; this point is called the origin. The position of any point P in the plane is then completely determined if its distance and direction from each of these axes is known. The units on the two scales are arbitrary ; they may or may not be equal to each other. The distance from either axis must, however, be measured in terms of the unit of the other axis, i.e. of the axis parallel to which the measurement takes place. Thus, in Fig. 20, where the units on the axes are different, the point Pj is determined by the distance a; = 3 units from the y-axis (meas- II, §34] ALGEBRAIC PRINCIPLES 39 ured in terms of the aj-unit) and the distance y = 2 units from the a>axis (measured in terms of the y-\mit). Similarly, the points P2, P3, P4 are determined respectively by the directed segments OM2 and M2P2, OM^ and M.^P^, OM^ and M^P^ ; the numbers representing these directed segments are signed numbers, so that the number gives both the magnitude and the direction of the segment. In such a system of rectangular coordinates in a plane, unless specifically agreed on otherwise, the positive direction on the a;-axis is always to the right; on the 2/-axis, always upward. We see, then, that every point in the plane is determined uniquely by a pair of numbers, and, conversely, that every pair of (real) numbers determines uniquely a point in the plane. The two numbers thus associated with any point in the plane are called the coordinates of the point ; the number X (giving the distance and direction from the ?/-axi^) is called the x-coordinate or the abscissa of the point, the number y (giving the distance and direction from the aj-axis) is called the y-cobrdinate or the ordinate of the point. Any point P in the plane may then be represented by a symbol {x, y), where the abscissa of P is written first in the symbol and the ordinate of P is written last. Thus we may write (Fig. 20) Pi =(3, 2) P, = (- 1, 4), P3 = (- V2, - 31), P, = (?, ?). The two axes divide the plane into four regions called quadrants, numbered as in the figure. The quadrant in which a point lies is completely determined by the signs of the coordinates of the point. Thus, the first quadrant is charac- terized by coordinates (-{-, +), the second quadrant by ( — , +), the third by ( — , — ), and the fourth by (+, — ). 34. Relations between Numbers. If two numbers a and b represent two points A and B respectively on an algebraic scale, we say that a is less than b (in symbols, a <b), it a is to 40 MATHEMATICAL ANALYSIS [II, § 34 the left of b, the scale being horizontal and the positive direction being to the right.* The following obvious relations are fundamental : (1) If a =5t 6, then either a < &, or 6 < a. (2) If a < 5 and 6<c, then a<c. EXERCISES 1. Is the date 1916 a signed number ? (Does it represent simply a duration of time or does it represent a time after some arbitrary fixed time ?) Would it be proper to represent the year 50 a.d. by + 50 and the year 50 b.c. by — 50 ? 2. When we designate the time of day as " two o'clock," is " two" a signed number ? 3. Are the (unsigned) integers used for any other purposes than to express the result of counting or measuring ? ( House numbers, catalog numbers, •••) 4. State some theorems of geometry concerning ratios. 5. Find a rational approximation of VS accurate to within 0.001. 6. Why is any actual measurement necessarily an approximation ? 7. Why is it incorrect to define a rational number as one " which does not contain radicals ? " 8. Why should irrational numbers be used at all, if it is possible to represent any such number by a rational number to as high a degree of approximation as may be desired ? 9. Explain /rom the definition of ratio why | in. and f^ in. represent the same magnitude. Why m/n in. and pm/pn in. represent the same magnitude, 10. Two segments measure | in, and f in., respectively. Show that the ratio of the first to the second according to the definition is y^^. (Ob- serve that ^ in. is a common measure of the two segments.) 11. Two segments measure m/n and p/q in. respectively. Prove that the ratio of the first to the second is mq/np. (Find a common measure of the two segments.) 12. Given that |a|<|6|, can we conclude that a<6? Why? Given that | a | > | 6 1, can we conclude that a > & ? Why ? * Likewise, a is greater than b (in symbols, a > b), if ^ is to the right of B. Obviously, if a < 6, then 6 > a. II, § 35j ALGEBRAIC PRINCIPLES 41 13. Which is the greater, -3 or -4? —S-lor — tr? 14. Locate on a line the points whose coordinates are 2, — |, |, — 2, 6. What is the distance between the last two ? What signed number represents the directed segment from the point + 5 to the point — 2 ? 15. Locate in a plane the points (2, 3), (— 2, 3), (2, — 3), (—2,-3), referred to a system of rectangular coordinates, the units on the two axes bfiing equal. 16. If the abscissa of a point is positive and its ordinate is negative, in what quadrant is the point ? If abscissa and ordinate are both negative ? 17. If the abscissa of a point in a plane is + 2, where is the point ? If the ordinate is zero ? What characterizes the coordinates of a point on the y-axis ? On the x-axis ? What are the coordinates of the origin ? 18. The units on the two scales being equal, what is the distance of the point (3, 4) from the origin ? Of the point (— 1, 7) ? Of the point (2, - 1) ? Of the point (a, b) ? 35. The Fundamental Operations. We shall now take up briefly the fundamental operations of addition, multiplication, subtraction, and division, and develop certain geometric inter- pretations and applications connected with these operations, which are of importance in what follows. Addition. We note first that the operation of addition for signed numbers has an essentially different meaning from that for unsigned numbers. The addition of two unsigned numbers expresses simply the addition of magnitudes. Thus, any two magnitudes may be represented geometrically by the lengths of two line segments. The segment, whose length represents their sum, is obtained by simply placing the segments end to end to form a single segment. (Compare the process of graphic addition described in Ex. 3, p. 7.) A signed number, on the other hand, represents a direction as well as a magnitude ; it is represented geometrically by a directed segment. Consider two signed numbers a and b. They will be represented by two directed segments whose 42 MATHEMATICAL ANALYSIS [II, § 35 lengths are | a| and 1 5 1, respectively, and whose directions are the same or opposite according as the numbers have the same or opposite signs. Figure 21 represents the four possible cases. The sum a + 6 is represented by a directed segment which expresses the net result of moving in the direction represented a y 1 b I 1 a + b 1 a\ 1 1 b ' ! ' i — ds: a+F o,+b Fig. 21 by a through a distance equal to | « |, and then moving in the direction of b through a distance equal to \b\. The segment representing a -f 6 is the segment from the initial point of these motions to the terminal point. (See Fig. 21.) The difference in the meaning of addition in the case of unsigned and signed numbers is clearly brought out by considering a simple concrete example : Suppose you walk to a place five miles distant and back again . The total distance you have walked is 5 + 5 = 10 miles. These are un- signed numbers. On the other hand, if you represent the trip out by + 6 and the trip back by — 5, which is entirely proper, the sum (+5)-f-(— 5), which is equal to 0, does not represent the distance walked at all, but does represent the net result of your walk measured from your starting point. The total distance walked is represented by | + 5 | -f- | — 6 |. It should be noted that the absolute value of the sum of two numbers is not, in general, equal to the sum of their absolute values. In fact all we can say in general on this point is that (1) k + ?>|<|«| + |^|.* The equality sign holds only when a and h have the same sign. * The symbol "^ is read " is equal to or less than." II, § 35] ALGEBRAIC PRINCIPLES 43 The geometric interpretations on the algebraic scale of add- ing a number x to all the numbers of the scale consists of sliding the whole scale to the right or left, according as x is positive or negative, through a distance equal to | a;|. Figure 22 illustrates this operation for the value x = — 2. Every number in the upper scale is the result of adding — 2 -4 -3 -2 -1 +1 +2 +3 '4 -3-2 -1 + i +2+3 +4 Fig. 22 to the number below it in the lower scale. Two important consequences follow from this interpretation : (1) If a <C b and x is any (real) number, then a -{- x <. b -\- x. (2) If a point P ivhose coordinate on a line is x is moved on the line through a distance and in a direction giveyi by the number h, the coordinate x' of its new position is given by the relation (2) x' = X + /i. An immediate consequence of the meaning of addition in the case of directed segments is as follows. If A, B, C are any three points on a line, then (3) AB -\-BC = AC, This relation holds no matter what the order of the pomts on the line may be. In fact it is obvious that to move on a line from Ato B and then to move from B to (7 is equivalent to moving directly from A to C, no matter how the points are situated on the line. As a special case of this relation we have AB-\-BA = 0, or AB = -BA. 44 MATHEMATICAL ANALYSIS [II, § 35 Multiplication. The product ab of two signed numbers a and b is defined as follows : (1) |a*| = |a|.16|. (2) The sign of ab is positive or negative according as the signs of a and b are the same or opposite. The statement (2) involves the familiar law of signs : (+)(+)=(+), (+)(-)=(-)(+)=(-). (-)(-)=(+)• Geometrically, multiplication by a positive number x is equivalent to a uniform expansion or contraction of the scale away from or toward the origin in the ratio | « | : 1, according as I a; I is greater than or less than 1. This statement will become clear on inspection of the follow- ing figure (Fig. 23) which gives the construction for the multi- FiG. 23 plication of every number on the scale by x. In the first figure X has been taken equal to + 2, in the second equal to -\- ^. The geometric interpretation of multiplication by a negative number x consist? of a similar expansion or co7itraction in the ratio I ic 1 : 1 combined with a rotation of the whole scale about the origin through an angle of 180°. For such a rotation will change each positive number into the corresponding nega- tive number, and vice versa, which the law of signs requires. Here again we may note two consequences of importance : 1. If a < b and x is any (real) number, ax is less than, equal to, or greater than bx, according as x is positive, zero, or negative. 2. If a scale is uniformly stretched (or contrax^ted), the origin II, § 35] ALGEBRAIC PRINCIPLES 45 remaining fixed, in sugJi a way that the point 1 moves to the point whose coordinate is a, then the point whose coordinate is x will mx)ve to the point whose coordinate is (4) X' = ax. Subtraction. To subtract a number b from a number a means to find a number x such that x-{-b = a. We then write x = a — b. Such a number x can always be found. Representing a and b by directed segments having the same initial point, the meaning of addition tells us at once that the segment from the terminal point of < — -- b to the terminal point of a represents the number x sought. (See Fig. 24.) ^'°- ^ This shows, moreover, that to subtract a number b is equivalent to adding the number — 5.* Division. To divide a number a by a number b means to find a number x such that bx = a. We then write x = a/b. It is always possible to find such a number x, except when the divisor b is zero. For we need merely reverse the construction given for multiplication (Fig. 23) as indicated in Fig. 25, first drawing the line joining b on the original scale to the point a on the multi- i 6\ plied scale and locating the required Fig. 25 point x on the multiplied scale by a line through 1 on the original scale, parallel to the line ab. In particular, we can always find a number x such tha! * It may be of interest to recall here the fact that historically the negative numbers were introduced in order to make the operation of subtraction always possible (i.e. even in the case when the subtrahend is greater than the minuend). But from what has just beeu said it appears that the device adopted for rendering the operation of subtraction more useful and convenient had the additional effect of making this operation unnecessary. 46 MATHEMATICAL ANALYSIS [II, § 35 bx = l,ifb=^ 0. This number 1/6 is called the reciprocal of h. Hence, to divide by b (b =^ 0) is equivalent to multiplying by 1/6. The Case 6 = 0. This case demands careful attention. Since • « = for every number x, it follows that the relation • x = a cannot be satisfied by any value of x, unless a is also ; and ivill be satisfied by every value of x, if a is 0. Hence, by the definition of division, the indicated quotient x-^ has no meaning whatever when a ^0, and no definite mean- ing even when a = 0. Hence, we conclude that division by zero, being either impossible or useless, is excluded from the legitimate operations of arithmetic and algebra. 36. The Function a/x. The Symbol oo . Whereas we have just seen that division by zero is not a legitimate operation, it is highly important for us to note what happens to the fraction a/x when x assumes values approachiyig nearer and nearer to zero ; as long as x does not equal zero, the indicated division is possible. We wish then to consider the /;mcfio?i a/x = y for values of x near 0. A table of corresponding values of x and y is as follows : X 4 3 2 1 \ \ -4 -3 -2 - 1 -I -I --: ia i« ha a 2a 4.a -\a -i« -la — a -2a -4a Plotting the points (x, a/x) with reference to two rectangular axes we obtain Fig. 26, where we have assumed a to be posi- tive and have chosen the unit on the oj-axis to be a times the unit on the 2/-axis. An inspection of the table and the graph shows us that as x decreases in absolute value, a/x increases in absolute value; II, § 36] ALGEBRAIC PRINCIPLES 47 more precisely, by choosing x sufficiently small in absolute value, a/x can be made as large in absolute value as we please. Further, when x = the expression a/x has no meaning. We say the fmicUon is not defined for the value x = 0; or, the range of the variable of this function does not include the value X Fig. 26 The sentence expressed in black-faced italics above is some- times written in a species of shorthand : 00. This looks like an equality involving a division by 0. But it does not mean any such thing. The expression a/0 as indi- cating a division by has already been pronounced illegiti- mate. For this very reason we are at liberty to use the symbol to mean something else without danger of confusion. We accordingly use it as a short way of expressing the values of the variable a/x as x is supposed to approach 0. Similarly, 48 MATHEMATICAL ANALYSIS [II, § 36 the symbol oo, read "infinity," does not represent a number at all, but a variable which increases loithout limit. The above equality is, therefore, an equality between variables, and is simply a short way of writing the phrase " as the denomina- tor of a fraction, whose numerator is constant and different from zero, approaches zero, the value of the fraction increases without limit in absolute value." Under these circumstances, we also say " the fraction becomes infinite." The phrase " equals infinity," which is sometimes heard, is very mislead- ing and its use should be strictly avoided. Returning to our table and graph, we note also that by assigning to x a value sufficiently large in absolute value, the value of a/x can be made in absolute value as small as we please, but not zero. The shorthand expression of this fact is 00 or " as the denominator becomes infinite the fraction ap- proaches 0." 37. The Directed Segment P^Pz- As an application of the foregoing principles we will now derive a formula which will often be used in the future. Let Pi and P^ be any two points on an algebraic scale, and let their coordinates be x^ and X2, respectively. We desire to find the number repre- senting the directed segment P1P2 in direction and magnitude. By definition x^ = OPi, Xo = OP2 (Fig. 27). Now, by § 35, Eq. 3, we have P1P2 = P,0 4- OP2 = - OPi + OP2 = -Xi + X2, or, finally, /^i* 2 ^ •^2 — ^1* Thus, if Xi = 2 and X2 = 5, X2 — x^ = -\- 3, and we conclude P2 P, Xi Q Xi Fig. 27 II, § 38] ALGEBRAIC PRINCIPLES 49 that the length of the segment P1P2 is 3 units and that its direction is positive (i.e. from left to right in the ordinary setting). On the other hand, if cci = 3 and Xg = — 4, we have X2— X1 — —I, and we conclude that the length of the segment is 7 and its direction is negative (i.e. P2 is to the left of Pj). 38. Concrete Illustration of the Law of Signs. The law of signs, as indeed many of the fundamental laws of algebra, is essentially a definition, arbitrary from a logical point of view and dictated largely on the grounds of convenience. The following concrete example will show how in one instance the conventions adopted in the law of signs for multiplication correspond to the concrete facts to be described. If a train moves at a constant speed of v miles per hour, then in t hours it will travel a distance = ^?i miles. Here v, t, s are unsigned num- bers. Now, let us change the formulation somewhat, so as to introduce the direction. At a given instant let the train be at a certain station 0; let us count time from this instant (t = 0) so that any positive t desig- nates an instant a certain number of hours after the instant t = 0, and a negative t designates an instant a certain number of hours before t = 0. Further, let the position of the train be determined by the signed number s representing the distance and the ^^ direction of the train from 0, s being positive [ ^ ^ if the train is to the right of (Fig. 28). |=o *"■*" Finally, let the speed and the direction in -p^^ 28 which the train is moving be given by the signed number v, v being positive if the train is moving to the right (u = — 30, for example, would mean that the train is moving to the left at the rate of 30 miles per hour). Now consider the four cases : (1) v and t both positive ; (2) v positive and t negative ; (3) v negative and t positive ; (4) v and t both negative. Verify that the law of signs in the relation s = vt gives the sign to s for which the actual position of the train in each case calls. [For example : (1) If v and t are both positive, s = vt will be positive, which is as it should be. For if the train is moving to the right, then a certain number of hours after t = 0, when the train was at s = 0, it will be a certain number of miles to the right of 0. (2) If v is positive and t negative, s = vt is negative. This also is correct. For a train moving to the right and arriving at O when t = 0, was to the left of at any time before t = 0. Etc. ] 50 MATHEMATICAL ANALYSIS [II, § 38 EXERCISES 1. Under what conditions is|a + &| = \a\ +\b\? 2. Prove that if A, B, C, D, •••, i, M are any points on a line (in any order) then AB + BC + CD + •" + LM = AM. 3. Graphic Addition. Given the directed segments, a, 6, c, d, e on parallel lines (or on the same line), their sum a+6+c+cZ + e may be found graphically as follows: On the straight edge of a piece of paper mark a point ; lay the strip along the segment «, the point coincid- ^ - ing with the initial point of a ; mark the ter- '' *■ minal point of a on the paper. Then slide the paper parallel to itself so as to make it lie along h and bring the mark just made into coincidence with the initial point of h ; mark the end-point of h. Then proceed similarly for the segments c, d, e. The directed seg- ment from to the final mark will then represent the sum sought. Why ? 4. Draw directed segments representing the numbers — 3, + 5, + 2, — 6, and find their sum graphically. 5. Show how to construct a directed segment representing the prod- uct of the numbers represented by segments a and h. [Hint. Use the adjoined figure to determine the magnitude of the product ; then determine the direc- tion. Observe that for the construction of a product we need to know the length of the unit segment, which was not necessary for a sum.] 6. Show how to construct a segment representing a/h. 7. Determine the numbers representing the directed segments from the first point of each of the following pairs of points to the second: -f- 8 and + 6, + 8 and -6,-2 and - 4, — ^ and + |, + 1.4 and - 2.1, — | and - I, + -V- and + 3.14. 8. By computing the numbers representing the segments, verify the relation AB + BC = AC, when the coordinates of A, B, C are, respec- tively : (a) 2, 3, 4 ; (6) 2, - 3, 4; (c) - 2, 3, - 4 ; (d) -2,-3, 4. 9. Find the coordinate of the mid-point of the segment joining the points whose coordinates on a line are 4 and 8 ; — 3 and 5; — 2 and — 5 ; Xi and X2. II, § 40] ALGEBRAIC PRINCIPLES 51 39. Insight and Technique. Most of our activities involve two more or less distinct aspects : insight and technique. On the one hand, we need to understand the nature of the thing we are trying to do, on the other we need skill in doing it. Theory and practice, planning and carrying out the plans, etc., are other ways of pointing the same distinction. In your previous study of arithmetic and algebra the major emphasis was on the side of technique. You learned at that time how to carry out certain manipulations with numbers ; and you gained more or less skill in using the processes. In the present course, the emphasis is to be placed on the side of in- sight, understanding, appreciation ; the technique of algebra is to be used merely as a tool, not as an end in itself.* 40. Definitions. We propose now to recall very briefly a few of the more important conceptions and processes of algebraic technique. We shall begin with the definitions of a few terms. When two or more numbers are added to form a surriy each of the numbers is called a term of the sum. When two or more numbers are multiplied to form a 2)roduct each of the numbers is called a factor of the product. Any combination of figures, letters, and other symbols, which represents a number, is called an expression. If the equality sign (=) is placed between two expressions, the result is called an equality, and the two expressions are called the members or the sides of the equality. An equality states that the two expressions represent the same number. -f * However, we must maintain a certain amount of proficiency in the use of algebraic processes. Hence " drill exercises " will not be wholly lacking in what follows. t Such a statement may or may not be a true statement. See § 47. 52 MATHEMATICAL ANALYSIS [II, § 40 Thus, suppose a, b, c, d, p, x, y represent numbers. Then a — hx -{- 1 cdy = a (12 y"^ — p) is an equality. The left-hand member is a sum of three terms; the right-hand member consists of only one term, which is a product of two factors. The second term of the left-hand side is a product of two factors, while the second factor of the right- hand side is a sum of two terms. 41. General Laws of Addition and Multiplication. The following general laws we take for granted : I. Concerning Addition : 1. Ariy tivo numbers may be added and their sum is a definite member. 2. The terms of any sum may be rearranged and grouped in any way without changing the sum. Thus, if a, &, c, p, q represent any numbers whatever, we have, for example, a -{-{b + c -^p) -\- q ={b -{- q) -\- {c -\- a) -\- p. II. Concerning Multiplication : 1. Any two numbers may be nudtiplied and their product is a definite number. 2. The factors of any product may be rearraiujed and grouped in any way without changing the product. Thus, if a, 6, c, x, y represent any numbers whatever, we have, for example, {abc){axy)= a^bcxy ={yx){cba'^). III. The Distributive Law : To multiply any sum by any number m, we may multiply each term of the sum by m and add the resulting j^foducts. Thus, (a -{- b -\- cd -\- — -\- x)m = am -}- bm -f cdm + — + xm. lY. The Law of Factoring : If every term of a sum con- tains the same number m as a factor , the sum contains m as a factor. Thus am -\-bm -^ cdm -f ••• -f xm = m (a -f- & + cd + — -f x). 11. § 43] ALGEBRAIC PRINCIPLES 53 Observe that IV is obtained from III by simply interchang- ing the sides of the equality. 42. Raising to Powers. Integral Exponents. We recall also at this point the meaning and use of integral exponents. The symbol cc", where x represents any number and n is any positive integer, is an abbreviation for the product of n factors each equal to x, i.e. x"" = x- X'X-" to n factors. From this definition and Principle II (§ 41) it follows at once that /pm^n =(x'X' X '" to m factors) (x • x 'X ••• to n factors) = X ' X • X "• to m + n factors, and therefore Ya x»^x^ = x^^'\ Similarly V6 — = x^^'^, if m > n, Xn Xn ;t" if n > m. (xmy = x"" • x"" ' x"^ •" to n factors r^m+m+m+ ■•■ ton term" and Also and therefore VI Also Vila VII6 43. Axioms. Closely connected with Principles I, 1 and II, 1 are the familiar axioms VIII a If a = h and c = d, then a -^ c = b -^ d. VIII b Ifa = b and g = d, then ac = bd. (^m)n = x**'^^. = a»b*K 64 MATHEMATICAL ANALYSIS [II, § 43 EXERCISES 1. Distinguish between insight and technique in the various professions (surgery, dentistry, engineering, etc.). 2. Complete the following propositions : (a) The sum of any two integers is • • • (6) The product of any two integers is • • • 3. What is the familiar expression in words for Principle VIII ? 4. Find the results of the following indicated operations : (1) a;i0a;i2. (6) x^^-^x^ (11) (av^y. (2) a^a^. 0) a^-^a^^. (12) (- cM»y. (3) ft^fes. (8) -^. (13) (-0*. a" (4) y2ny3«. ^g-j ^^4^8. (14) (r^S-)". (5) X'»-%2. (10) (C2)6. ' (15) (-X2)». 5. Multiply x^* + x^y^ + i/2& by x2» — x'^y^ + y"^. 6. Divide x^" + y^n by x" + ?/". 7. Perform the following operations : (1)25.24= (2)25.44= (3)32.23= (4)7i5^7i3 = 44. Discussion of Principles. In the preceding article Principles V-VII were derived from I and II, while IV is a consequence of III. We might now ask : " How do we know that Principles I, II, and III are true for all numbers ? " On these three principles the whole subject of algebraic technique rests. They are so simple that they may appear at first sight to be trivial. As a matter of fact their truth is by no means obvious ; our unquestioned belief in them is the re- sult of experience in using numbers. Were we to attempt a general proof, we should find it a long and difficult process which is out of place in an introductory course. Hence we simply take them for granted. A little reflection will show that these principles are not obvious. Take for example the fact implied by Principle II, 2 : a times b is equal to h times a ; and let us suppose that a and b are positive integers. Now, 2x3 means 3 + 3 and 3x2 means 2 + 2 + 2. By addition we observe II, § 44] ALGEBRAIC PRINCIPLES 55 that the result is in both cases 6. But that simply verifies the general law when a = 2 and 6 = 3. We can thus verify the law in question for any two special integral values of a and h. Not only would this be ex- tremely laborious for large values ; it would still be only a verification for a special case ; it would not be a general proof. Moreover, we have con- fined ourselves to the simplest of all numbers, the positive integers ; while II, 2 asserts among other things that ah = 6a, no matter what numbers a and 6 represent (rational or irrational, real or imaginary) . As has been indicated in the preceding paragraph, we are not concerned in this course with proving these principles. Of great interest to us, however, are the relations existing between numbers and geometry. Accordingly we have suggested in the exercises below some geometrical interpretations of these principles. which furnish intuitive proofs of certain restricted cases. EXERCISES 1. An intuitive proof that ah = ha, in case a and 6 are positive inte- gers : Let the integer a be represented by a group of a equal squares placed side by side so as to form a row (see the figure, where a = 8). The product 6 • a is then represented by the number of squares in 6 such rows. Show that, if these rows be placed under each other (as in the figure, where 6 = 6), it is seen that the total number of squares is also equal to the number of squares in a columns each containing 6 squares. Observe that while the figure is drawn for special values of a and 6, the argument is general. 2. From a consideration of the adjacent figure give an intuitive proof that 5 • (3 • 4) = 3 • (5 • 4) . Then by using the fact that ah = ha show that (3 • 4) • 5 = 3 • (4 . 5). Can this argument be made general to show (a • 6) • c = a • (6 • c), when a, 6, c are positive integers ? 4 4 A 4 4 4 4 4 4 4 4 4 .4 4 4 66 MATHEMATICAL ANALYSIS [II, § 44 3. From the adjacent figure, show how to use the idea of a rectangu- lar pile of blocks to prove that (a • 6) • c = a • (& • c), when a, &, c are positive integers. ^--^^4^-^ 7^ ^ •^ j^'O^O^O' >' ^ u^ 4. Assuming that the area of a rectangle is equal to the product of its base by its altitude, show that ab — ba, when a, b are any positive real numbers. 5. By considering the adjacent figure, interpret a b geometrically the relation {^a ■\- b)c := ac -{■ be. 6. Interpret geometrically the equalities (a) (a + &)2 = a2 + 2 a6 + &2. (6) (a + 6) (c + (?)=: ac + he ^- ad ^- bd. d € 7. Derive the equalities in Ex. 6 from Principles I-V. [For 6 (6), we first consider c + d as a single number. Ill then gives (a + 6)(c + d) = a(c + d) + 6 (c + d). Applying II and III to each of the terms of the right-hand member of this equality, we get the desired result. ] 8. Show in detail how the carrying out of the product (ax+6)(cx+d) involves Principles I-V. 9. Show how these principles apply in the addition of 2 x^ + 7, 3 x + 2, and 4 x2 + X -}- 3. II, § 45] ALGEBRAIC PRINCIPLES 57 45. Review of Algebraic Technique. We propose now to take up a few of the most elementary portions of the tech- nique of algebra. These are all that will be needed in the immediate future. Other topics relating to technique will be recalled when they are needed. The technique of algebra is concerned altogether with the transforming of expressions into other equivalent expressions which serve better the purpose in hand. The principal pro- cesses used are the following : (a) Performing indicated operations and collecting terms. For ex^^ ample, collect the terms in x, ?/, and z in the following : --^ 2x4-72/ -3^; + y -\x—^y ^bz^ 3x. T"..''/' The result \%x -\-1z. This involves Principles I and IV. Perform the indicated operation and collect terms in (x2_3a;-|-4)(x-2). The result is a;-^ — 5 a;^ + 10 a: — 8. This involves Principles I-V. (b) The use of special products. The following equalities should be memorized : (1) (a-}-b)(a-6)=a2-62. (2) (a + b)2 = a^ + 2a& + &2. (3) (a-&)2 = a2-2ab + 62. (4) (AC + a)(x -h b)= x2 + (a + 6) X + ab. (c) Factoring. The following cases may be specially mentioned : i. I%e difference of two squares. Use special product (1). Thus ^9x^ - iy^ ={7 x« + 2y)(7 x^ - 2y). ii. Trinomials of the form x^ + px + q. Try to find two numbers whose sum is p and whose product is q, in accordance with special product (4) . Thus to factor x2 _ 6 X - 27 we notice that 3 and — 9 are two numbers which satisfy the requirements. Hence, a;2_6x-27 = (x+ 3)(x-9). Again, to factor x2 — 10 x + 25, we notice that — 5 and — 5 are two numbers satisfying the required conditions. Hence x2 _ lOx + 25 = (x - 5) (x - 5) = (X - 5)2. 58 MATHEMATICAL ANALYSIS [II, § 45 EXERCISES 1. Perform the following operations : (a) (a; -4) (2 a: + 3). (&) (x + a)(x + b)(x + c)' (c) (x + h)^-x^. (d) (m + w) n — (m — n)m. (e) (a + &)2-(a-6)2. (/) (a + 6)3 -(a -6)3. 2. Factor: (a) x^-W. (&) (2a-by-9(x-iy^. (c) ax — bx + ay — by. (d) x*- 6x^ + 9. (e) x2 + 6x + 6. (;i) 25x* + 10a;3 + x^. 3. Factor: (a) 4a2_5« + i. (6) a2 + 2a6 + 62_a;2. (c) a9-64a3_a6 + 64. ((^) 1 + x2 + xK (e) x3 -3a;2 + 2a;. (/;2 + 7x-15x2. (Sr) x8 + 1. (/i) a:4?/2 _ 17 x'^y - 110. 46. Operations with Fractions. These depend chiefly on the simple principle that the numerator and the denominator of a fraction may be multiplied by the same number (not zero) without changing the value of the fraction, and the reverse of this principle, viz. that any common factor (not zero) of the numerator and the denominator of a fraction may be removed without changing the value of the fraction.* By means of this rule any two or more fractions can be re- duced to the same denominator. The rules for adding and multiplying fractional expressions are stated symbolically as follows : a b +r b c a b xr ac 'bd a b~ na ''nb * This principle is a direct consequence of the definition of division. Can you explain it ? II, § 46] ALGEBRAIC PRINCIPLES 59 IBR^ iK : PRINCIPLES a b +r ad-\- be bd a b_ a "b d • — — c ad = — . be d a b = - — a b a -b a + &_ -a-b Also and c c The following exercises will furnish applications of these principles. EXERCISES 1. Express as a single fraction : ^^^ M^ + ^T^ + ^T&' ^ ^ c^y a'^ b^^ ^ X y z x + y -\-z \n vj\m uj (c) L^l + L. (^) 1 + 1_§. , St tr rs ^ 2 X y Simplify the following expressions, assuming that no canceled factors have the value zero. a-\-b 2. X - -— + - a {X- -aY \x -ay 3. a - - c h + c (a-b){c-b) {a-c)(a-b) {b-c){c-a) Ans. 2« (a_6)(c-6) «J=L^ _ 1±A I (62 _ an Ans. 4 ab. a + b a-&P ^ g x^ -y* ^ g-i^ ab + b^ ' \a^ b^\ ' \a^ b^ 60 MATHEMATICAL ANALYSIS [II, § 46 be ac ab a) ( i _ ^ c V a + & + c \a^-^b^)\ b I \b a) ^2 ^. 7y2 q;2 _ ft2 ) 10. 11. aP- - b-^ a' + b-^ a + b a-b 1 a-b a + b ^_^x^ 2\x-y x-\-y) 1+-^+ . "".J M- xy + 4 y'^ x^ + 4:xy y(x + 2 y) Aiis. a 111 c U (a + 6)^-c° l I a + 6|l (a + 6/ j i + l + i 13. 2_L_2. £ + « + * a & c 14 a^ + &'^ ^ ■ ?) \ g^-fe^ a^ + rt6 + ^n a- bl a^-ab + b^' X^ + -J^\ {X'^ + y^} X'^ 1/2 '- 15. -^ ^-^ Ans. X* - a;2y2 + y*, ^ I y x-hy x — y 16. If a, ft, and « are positive, which is the greater, or b (b-{-x) Distinguish two cases a>- b^ a <.b. 17. If ad < eft, then is it true for all values of the letters involved that a/b < c/d ? Why ? II, §48] ALGEBRAIC PRINCIPLES 61 47. Identities and Equations. We must recall here a vital distinction between two kinds of equalities. An equality which is true for all values of the letters (or other symbols) involved, for which both members of the equality have a mean- ing, is called an unconditional equality or an identity. An equality which may be true for certain values of the letters involved, but is not true for all, is called a conditional equality or an equation. For example, the equality a'^ — b^=(a— b) (a + b) is an identity since the two members of the equality represent the same number for all values of a and b. Also the equality ^3^ =^ + ^ is an identity, even though it becomes meaningless when a = b. Why ? On the other hand 2^ — 8 = is an equation since it is true only for x = 4. Va? + 1 = — 1 is also an equation, but it is not true for any value of x. Why ? * To solve an equation is to find the values of the letters for which it is true. Thus in the first example above, a; = 4 is the solution or root of the equation 2 a? — 8 = 0. The second equa- tion above has no root. 48. The Relation AB = 0. In the solution of equations the following principle is of frequent application. If a prod- uct of expressions each representing a number is zero, we may conclude that some one of the factors is zero. In the simplest case this means that if A and B represent expressions and if A • B = Oy we conclude that either ^1 = or J5 = O.f * By Va is meant the positive square root of a. t We must, of course, be careful to assure ourselves that each of the expres- sions involved represents a number for the values under consideration. Thus we cannot conclude from the relation x • (l/x) = 0, that either a = or 1/x = 0, for when a; = 0, l/x is meaningless. In fact the given relation is impossible; the equality is not true for any value of x. 62 MATHEMATICAL ANALYSIS [II, § 48 We may apply this principle to show the absurdity of some mistakes that are often made by the careless student. For example, a favorite mistake is to " cancel " the x in the expres- sion a-\-x This would be justified if the equality a\ a + x ^a ^^ b+x b were an identity. If we clear this equality of fractions by multiplying both members by (b + x) b, we obtain ba '\- bx = ab -{- ax, or bx = ax; or, finally, (b-a)x = 0. Hence we conclude that equality (4) cannot be true, unless either 6 = a, or ic = 0. The " canceling operation " mentioned above is therefore unjustified. EXERCISES 1. Treat similarly the following equalities to determine under what conditions they are true. Each one is related to an error that is some- times made. (a) Is Va2 + 6'-^ = a + 6 ? (Square both sides.) ^ ^ b d b-\-d ^ ^ 2c + d c + d (d) Is (x + 2/)2 = x2 + y2 ? (e) Given x^ = 2x. Are we justified in concluding that x = 2 ? II, § 48] ALGEBRAIC PRINCIPLES 63 2. In each of the following equalities, assuming that the letters repre- sent real numbers, determine which are identities and which are equations. Among the latter, distinguish those that are not true for any (real) values of the letters involved ; and for the others determine in their simplest form the conditions which they imply on the letters involved. (a) a;*-y4=(x + 2/)(x-y)(x2+|/2). (6) x'^-Sx + 2 = 0. (c) x+- = 0. X (d) ac — he + ad — bd = 0. 3. Find and discuss the error in the following reasoning : Let X = 2. Then a;2 = 2 x, and x^ — 4 = 2x — 4. This is equivalent to (a;+2)(x-2) =2(:b-2). Dividing both sides by x — 2, we get cc + 2 = 2. But X = 2 ; hence 2 + 2 = 2 or 4 = 2. 4. Find and discuss the error in the following reasoning : Let a and b represent two numbers. Then a2 _ 2 a6 + 62 =6-2 _ 2 a& + a% or (a-.6)2=(6_a)2, or a — b = b— a; hence a = 6. PART II. ELEMENTARY FUNCTIONS CHAPTER III THE LINEAR FUNCTION. THE STRAIGHT LINE 49. A Linear Function. Distance traversed at uniform speed. Example. A railroad train starts 10 miles east of Buffalo and travels east at the rate of 30 miles per hour. How far from Buffalo is the train at the end of x hours ? In oc hours the train travels 30 x miles. If its distance from Buffalo is denoted by y, we have 2/ = 30 a; + 10. Pairs of values of X and y obtained from this equation are shown in the fol- lowing table. X 1 2 3 4 etc. y 10 40 70 100 130 etc. Geometric Eepresentation. Let us plot as points in a plane these corresponding values of x and y. We then obtain the first of the following figures (Fig. 29). It will be noticed Y Y Y I ' • 3 4X i i ^ ^X 12 3 4 Fig. 29 64 Ill, §49] LINEAR FUNCTIONS 65 that the five points appear to lie on a straight line. We have, for intermediate values of x, the values of y shown in the fol- lowing table. X i ^ ■ 2i H y . 25 55 85 115 Plotting these points we obtain the second of the above figures, in which the nine points appear to lie on a straight line. Let us calculate the value of y for some more intermediate values of X thus : X JL 4 f i i y m ?>2i 47i 62^ In the third figure we see that these new points still appear to lie on the same straight line. These considerations suggest that if we could calculate the values of y corresponding to all the values of x between x = and a? = 4, the points whose coordinates are {x, y) would all lie on a straight line joining the points (0, 10) and (4, 130), and would constitute the whole of this line-segment. A proof that this is the case is as follows : In Fig. 30 we have drawn the straight line joining the points A (0, 10) and B (4, 130). Let (x^, y^)(x, > 0) be any pair of corresponding values of x and y for the function 2/ = 30 ic + 10 ; we then have (1) 2/1 = 30 0^1+10. . ,,, 120 -^ - J- «! i^ 80 - Al 60--'- :::^ :n ^'^ Jz t^ 40 - ^ ^ 1 ^ /^ II 20 ^ til- ' A 4 - yj, ^ ^ i-iS-i-^-SElL 1 2 M^ * Fig. 30 66 MATHEMATICAL ANALYSIS [III, § 49 We now wish to prove that the point P (a^i, y{) is on this line AB.* To do this we construct the triangles APQ and ABC by drawing lines through A, P, and B parallel to the axes. If P is on the line JLB, then these triangles are similar, and if P is not on the line AB, then the triangles are not similar. Why? If the triangles are similar, QP/CB = AQ/AC; and conversely, if QP/CB = AQ/AC, the two triangles are similar. Expressed in terms of the coordinates of A, By and P, this pro- portion becomes (see figure) 120 4 ' or ^\i~'' 3/1-10 -120^3^ Xi 4 But from (1) we have y^ — 10 = 30 Xi and hence ^-Zi? = 30. Xi This proves that every point whose coordinates {x^, y-^ satisfy the relation 2/ = 30 a; + 10 is on the straight line AB. Conversely, every point on the straight line AB has coordinates {xij yi) which satisfy the relation y = SO x -\- 10. For, from the figure, we have yi-10 ^120 Xi 4 whence yi=S0x^-\-10. * Extended beyond B, if Xi > 4. ** Observe that QP and CB are measured in different units from AQ and AC. But the ratio of two line-segments is independent of the unit in which they are measured. Ill, § 49] LINEAR FUNCTIONS 67 The straight line AB (extended indefinitely beyond B) then gives a complete representation of the function 30 a; + 10, at least for positive values of x (negative values of x have no meaning in this problem). Every pair of corresponding values of X and y gives rise to a point on AB, and eveiy point of AB has coordinates which are corresponding values of x and y. By virtue of this fact the line xiB is called the graph of the function 30 a; 4- 10, or the locus of the equation y = 30 x -{- 10 referred to rectangular coordinates ; Avhereas the equation ?/ = 30 a; + 10 is called the equation of the line AB. Uses of the Graph. The graph just discussed exhibits vividly to the eye several properties of the function 30 a; 4- 10. (1) The function steadily increases as x increases. This corresponds to the fact that the longer the train moves east- ward, the greater is its distance from Buffalo. (2) Corresponding to every positive value of a;, there is a unique value of y. From the graph find y when x is 4. (3) Corresponding to every positive value of y (greater than 10) there is a unique value of x. What is the value of x when .vis 160? (4) The last consideration means that x is also a function of y. Explicitly we have y = 30 a; + 10, whence 2/ - 10 = 30 a; and x=.y^^. 30 •It is left as an exercise to draw the graph of the function y-10 ^='"^0~ by assigning values to y and computing the corresponding values of x. Compare the result with the graph in Fig. 30. 68 MATHEMATICAL ANALYSIS [III, § 49 Rate of Change of a Functiox. Before leaving this special case to consider a more general problem, we shall use it to illus- trate a very important conception connected with a function. We have noted that when a; = 0, y = 10. Starting from this initial value, as x increases from the value 0, the value of the function, i.e. _?/, changes (in this case increases). It is often of the greatest importance to know how the increase in the func- tion y is related to the increase in x. As x increases from to 1, y increases from 10 to 40 ; i.e. a change in x of one unit produces a change in ?/ of 40 — 10 or 30 units. The relative change is then -^^2^, or 30. As x increases from x = to x = 2, y changes from y z= 10 to y= 70, or by 60 units, and the relative change is again 30. Let us see what the situation is in general. Let Xi be any particular value of x and yi the corresponding value of y ; then suppose that Xo is any other (subsequent) value of x and ?/2 the corresponding value of ?/. The change in x is evidently X2 — Xi and the corresponding change in y is y2 — 2/1- We seek the value of the ratio X^-Xi We have from the data of the problem 2/2 = 30 X2 + 10 and y^ = 30 a^i -h 10. Subtracting we get 2/2-2/1 = 30(^2 -aJi) and hence ^2-3/1^30, 0^2 — a?! We see then that the ratio of a change in the function 30 a; +10 to the corresponding change in x is constant and is equal to Ill, § 50] LINEAR FUNCTIONS 69 the speed of the train. We shall see presently that in any function of the first degree in x, the ratio of a change in the function to the corresponding change in X is constant. Geometrically this result expresses the familiar proportionality of homol- ogous sides of similar triangles. By reference to Fig. 31 we may readily verify that the terms 2/2 — 2/i ^^^^ ^2 — ^1 represent the vertical and the hori- zontal sides of a right triangle whose hypotenuse is on the line AB. The fact that the ratio (2/2 — 2/i)/(^2 — ^1) is constant, i.e. always equal to 30, simply corresponds to the obvious fact that any two such triangles, no matter at what place they are drawn, or how long their sides are taken, are similar. \ l^' IV inn ~ A t" ^^ V ^-- i yfr T^ eo ~" ,> 1:1 1 1 \\r\ 1 bU ^^ I ISI 40 -P -.^ 4^ : 20-i'^V Xn- = •^1--- + -- + ^ 2 3 i Fig. 31 50. Change Ratio. The ratio X2 — X1 is called the change ratio (or sometimes the difference ratio) of the function. The difference X2 — x^ is often denoted by Ace, and the corresponding difference y^ — 2/1 by Ai/.* The change ratio may then be written A?//Aiw. Explicitly, by definition, we have the following equalities : change ratio = ^ = fclli^ = change in y . Ajt X2 —Xi corresponding change m x The preceding considerations suggest the theorem : * A is a Greek capital letter corresponding to our D and called delta ; it is used because d is the initial of the word ** difference." " Ax," is then merely an abbreviation for "difference of the x's " or '* change in z " and " Ay " for " difference of the y's " or " change in y." 70 MATHEMATICAL ANALYSIS [III, § 50 If the change ratio of a function is constant, the graph of the function is a straight line; and conversely. The truth of this theorem is already sufficiently indicated in case 2/2 — Vi ^^d ^2 — ^1 3.re both positive numbers. In formu- lating a general proof we must keep in mind that y^ — 2/1 ^^d X2 — Xi may be either positive or negative and that these dif- ferences represent directed segments. The proof of the theorem in general will appear presently. EXERCISES 1. Discuss fully the graph of the function y = 2 a; + 3. Prove that the graph is a straight line. Express x as a function of y. Find the change ratio and show that it is constant. 2. Proceed as in Ex. 1 for each of the functions : (a) 5x4-2, {h)x+\2, (c) 3.2a: + 8.4. 3. Prove that the change ratio for the function y = mx + 6 is m. 4. A steamer 150 miles east of Toledo starts to travel west at a uniform rate of 15 miles per hour. Express its distance y east of Toledo at the end of X hours. Draw the graph of the function and prove that it is a straight line. Does the distance y increase as x increases ? Calculate the change ratio and show that It is constant. What is the significance of the negative sign ? At what time is the steamer 10 miles east of Toledo ? When does it reach Toledo ? How are the last two results shown in the graph ? What is the significance of the graph that extends below the x-axis ? 5. Give examples, drawn from your experience, of functions which (a) increase as the variable increases ; (6) decrease as the variable decreases. 6. Consider the function y ■= x^. Calculate the corresponding values of y when a; = 0, 1, 2, 3, 4, 5. Plot the corresponding points and observe that they are not on a straight line. Calculate the change ratio of this function for x = and Ax = 1, 2, 3, and observe that it is not constant. 7. The cost of printing certain circulars is computed according to the following rule. The cost for the first one hundred circulars is $ 2 and for each succeeding one hundred $0.50. Express the cost y in dollars of X hundred circulars. Draw the graph of the function and determine from the graph the cost of printing 475 circulars. What does the change ratio of the function y express in this case ? Ans. y = ^ x + f . Ill, § 51] LINEAR FUNCTIONS 71 51. The General Linear Function mx -h b. The change ratio of every function of the form mx + b is constant. Proof. Let (x^ , yi) and (ajg , 2/2) be any two pairs of cor- responding values. Then 2/1 = w^i + & and 2/2 = mx2 -\- h ; hence , ^ 2/2 - .Vi = m (icg - xi), I.e. Conversely, if the change ratio of the function y of x is con- stant and equal to m, the function has the form y = mx + b. Let (xi, 2/1) be a particular pair of corresponding values and (x, y) any other pair of corresponding values. By hypothesis the change ratio is equal to m ; i.e. X — Xi or y = mx — mxi -f 2/1 ; but — mxi 4- 2/1 is a constant, say b. Hence y = mx -h b. Hookers law affords an excellent illustration of the above theorem. This law states that the length 1/ of a piece of wire under tension is equal to its original length 6, plus the stretch, which is proportional to the force X causing it. Thus, y z=l h + mx. This law may also be stated simply by saying that the change ratio of the length y^ with respect to the pull a:, is constant. The preceding considerations lead to the following theorem. Theorem. If a function y of a variable x is such that any change in the value of the function is always equal to m times the corresponding change in the variable, the function y is given by a relatian of the form y — mx -f- b, and, conversely, in any func- tion of this form any change in y is always m times the corre- sponding change in x. 72 MATHEMATICAL ANALYSIS [III, § 52 52. The Graph of a Linear Function. Let Pi{xi , y^) be any point on the graph of a linear function (Fig. 32). From Pj draw to the right a positive horizontal segment P1Q2 equal in length to X2 — Xi , i.e. Ax. Through Q2 draw a vertical segment and let it meet the graph in the point Po • The segment Q2P2 is equal to 2/2 — yi? i'^- ^y, and is positive if P2 is above Pi (Fig. 32 a) }' . __ Pz Pi Q% Qs la) X Fig. 32 Pi Q2 Qb A A (6) and negative if P2 is below Pi (Fig. 32 6). Now let us take another positive change Ax = x^ — Xi, represented by P] Q.^ and the corresponding change Ay — y^ — 2/1 represented by Q3P3 . If the change ratio is constant, then (1) either Pg and P3 are both above Pi or they are both below Pj , according as the given constant is positive or negative; and (2) the triangles P1Q2P2 and P1Q3P3 are similar. Therefore the points P1P2P3 are on a straight line, if and only if the change ratio is constant. Theorem. The graph of any function of the form y = mx -f- b is a straight line. To draw the graph of such a function we need, therefore, merely to plot two points of the graph and draw the straight line through them.* * While two points are sufficient to determine the line completely, it is desirable to find a third point as a cheek on the other two. Moreover, it is advisable to take the points as far apart as convenient. Why ? Ill, § 54] LINEAR FUNCTIONS 73 Y C n B y^ D ,^ 1 y IM 2 X Fig. 33 In the figure this ratio Example. Draw the graph y = ^x -\- 2. We notice that (0, 2) and (4, 14) are two points on the graph. The line joining these two points is the required line. Check by plotting a third point. 53. The Slope of a Straight Line. The graph of the func- tion y = mx + b may be obtained by observing that x = Oj y -^b and x = 1, y = m -\- b are two pairs of corresponding values of X and y. In the adjoining figure (Fig. 33) we have plotted the two points B (.0, b) and 0(1, m + b) on the assumption that both of the quantities b and m are positive numbers. The change ratio, as we have seen, is m. is DC/BD. Now suppose that b remains constant and that m takes on successively different values. Under the hypothesis that b and BD remain fixed, the points B and D would remain fixed and the point C would move up or down on the vertical line through D, according as m increases or decreases. The line BG would then rotate about the point B, becoming steeper if m is increased and less steep if m is decreased. The change ratio m then measures the steepness of the line. The term change ratio applies to the function mx -\- b ; when applied to the straight line y = nix -f 6, it is called the slope of the straight line. 54. Remarks Concerning the Slope of a Line. We as- sumed in the last section that both b and m were positive numbers. Let us now suppose that b is still positive, but that m is negative. Observe that in the preceding figure MC = MD + DC = b-\-m. Eecalling that the relation MC= MD + DC holds universally for any three points M, D, C, on a 74 MATHEMATICAL ANALYSIS [III, § 54 line (Art. 35), the interpretation of a negative m, i.e. DC, is that the point C is below, the point D. (Cf. also § 52.) A negative value of m then merely causes the line to slope downward in going from left to right, while, as we have seen, a line with a positive m slopes upward. When m = 0, the line is parallel to the a^-axis. Indeed the equation y = mx + h be- comes, for the value m = 0, the equation y = h. This equation, when interpreted as a function of x, means that for every value of X, the value of ?/ is h \ the graph of such a function is ob- viously a straight line parallel to the avaxis. Since a change in X in this case produces no change in y, the change ratio is zero. Finally, if h is negative, nothing is changed except that the point B is below the origin 0. A positive m still indicates an upward slope and a negative m a downward slope, in pass- ing from left to right. The number h, we have seen, represents the segment from the origin to the point in which the line cuts the y-axis. This segment is called the y-intercept of the line. Similarly, the segment from the origin to the point in which the line cuts the ic-axis is called the x-intercept of the line. We have then the following results : Tlie straight line represented by the equation y = mx + b has a slope equal to m a7id a y-intercept equal to b. In passing from left to right, the straight line slopes downward ifm is negative and up- ward if m is positive; if m is zero, the line is parallel to the x-axis. In the terminology of functions we have : The linear function mx -f- b is an increasing function of x {i.e. the function increases as x increases) if the charige ratio m is pos- itive, and a decreasing function of x (i.e. the function decreases as x increases) if m is negative. It is a constant function if m is zero. Ill, § 55] LINEAR FUNCTIONS 75 55. Examples of Linear Functions. Example l. On a Fah- renheit thermometer the freezing point of water is placed at 32°, the boil- ing point at 212°. On a Centigrade thermometer the freezing point is at 0°, the boiling point at 100^. Express the temperature of y° Fahren- heit as a function of y° Centigrade. Solution : y = 32 when a; = 0. Also the range of temperature from the freezing point to the boiling point of water is 212°-32° or 180° F. while it is 100° C. Therefore it follows that an increase of 1° C. is equivalent to an increase of | of a degree F. Now as the temperature increases from 0° to ic° C. the change in the number of degrees is x. This change in temperature is equivalent to an increase from 32° to y° F. The change in the number of degrees is then ?/-32i=|x, or y =|x + 32. As a check we may observe that, when x = 100, the formula gives y = 212, as it should. Are negative values of x admissible ? Figure 34 represents the graph of this function. It was drawn by using the points A (-30, -22), B (100, 212). [Why is it desirable to choose points so far apart?] This "graph may be used to read off without computation the approximate temperature in F. for a given temperature in C. For example, to x = 22 corresponds y = 72, approximately. Therefore 22° C. is equivalent to about 72° F. By computation we find that y = ll.Q. ■ 1 ' ' " ' . .? zko i ^ ' / ^ t / '4 1 / js / ^ / *- ;5 1 Si) 2^ / y /\ / / / / / / / .-i 5 p'fl no 7 a . / I) e. rj. es 4 iti I' ii.(.e\ r / A 'A ._ .- 1 1 Fig. 34 76 MATHEMATICAL ANALYSIS [III, § 55 Example 2. A bar of iron 3 ft. long at 60"^ F. will expand or con- tract if tlie temperature increases or decreases. The increase in length is proportional to the increase in temperature (physical law). More pre- cisely, an increase of 1° F. produces an increase of 0.0000027 ft. In this case we have m = 0.0000027. If y represents the length at x° F., we have ?/ = 3 + m(x-60). Does this relation hold also when x < 60 ? Why ? Can you draw the graph ? We shall now give an example in which m is negative. Example 3, An aeroplane starts 200 miles eas^of Chicago and travels towards Chicago. Express its distance y from Chicago in miles at the end of t hours, if the aeroplane moves at the rate of 82 miles per hour. Solution : According to the data the distance from Chicago is de- creasing at the rate of 82 miles per hour, i.e. m =— 82. Therefore, y-2Q0=-82t, or y = -S2 t -\- 200. Draw the graph. What is the significance of a negative y (e.g. when ^ = 5)? When does the aeroplane reach Chicago ? When is it at a point 63 miles east of Chicago ? When is it 52 miles west of Chicago ? How could these questions be answered from the graph alone ? EXERCISES 1. On a Reaumur thermometer the freezing point of water is at 0°, the boiling point at 80°. Express the temperature in degrees Fahrenheit in terms of the temperature in degrees Reaumur. Draw the graph and show how it may be used. 2. Is there any temperature whose measures in the Fahrenheit and in the Centigrade scales are equal ? Answer by computation. How could the result be found graphically ? 3. A cistern that already contains 300 gallons of water is filled at the rate of 50 gallons per hour. Show that the amount of water y in this cis- tern at the end of x hours is y = 50 x -}- 300. Draw the graph and discuss. How would the function be changed, if the cistern were being emptied at the rate of 50 gallons per hour ? 4. A tank contains 16 gallons of water. A faucet is opened which admits 4 gallons per minute. Express the amount, w, of water in the tank at the end of t minutes. Draw the graph. Do negative values of t have any significance ? When will the tank contain 37 gallons ? Ill, §56] LINEAR FUNCTIONS 77 5. A tank containing 37 gallons of gasolene is emptied at the rate of 5 gallons per minute. Express the amount of gasolene in the tank at the end of t minutes. Draw the graph. When will the tank be emptied ? For what range of values of t has the function any significance ? 6. On a certain date a man has 1 5 in the bank. At the end of every week he deposits $3. How much money has he in the bank at the end of X weeks ? Draw the graph of this function. How is the rate of in- crease shown in the graph ? 7. On a certain day a man has 1 100 in the bank. At the end of every week he draws out '$5. How much money has he in the bank at the end of x weeks ? Draw the graph of this function. How is the rate of decrease shown in the graph ? 8. In experiments with a pulley, the pull P in jjounds required to lift a load L in pounds, was found to be P = 0.15 L + 2. Plot this relation. How much is P when L is zero. How much is P when L is 10 lbs.? 9. If h represents the height in meters above sea level, and h the reading of a barometer in millimeters, it is known that b — k ■{■ hm, where k and m are constants. At a height of 110 meters above sea level the barometer reads 750 ; at a height of 770 meters it reads 695. What equation gives the relation between b and Ji ? Draw the graph of this equation and from the graph determine h when b = 680. 56. Linear Interpolation. The fact that the change Ay in a linear function y is proportional to the change Aa; in the variable X makes it possible to interpolate readily. For example, if we know that ?/ is a linear function of x, and that y = 432.50 when X = 32.0 and that y — 436.90 when x = 33.0, we can calculate mentally the value of y when x = 32.3. For we know that in this case Ay = 4.40 when Ax = 1.0 ; hence A?/ = 4.4 x 0.3 = 1.32 when Ail' = 0.3. Hence y is 433.82 when x = 32.3. This pro- cess is known as linear interpolation. Why would this process not apply directly to functions that are not linear ? EXERCISES Assuming that y is a linear function in each of the following cases compute the values of y indicated. 1. When a; = 10, ?/ = 50; when x = 14, y = 90; when a; = 11, y = ? 2. When x = 2.4, ?/ = 9.8 ; when x = 2.5, y = 8.Q; when x = 2.42, y=? 78 MATHEMATICAL ANALYSIS [III, § 57 57. Graphic Solution of Problems. Whenever we know at the outset that the solution of a problem is going to depend on the consideration of one or more linear functions, we can often solve the problem graphically without determining these linear functions analytically. Such a method is advantageous whenever the computation is difficult or tedious and when great accuracy is unnecessary. In order to decide whether the functions involved are linear or not, we usually have re- course to the theorem (§ 51) that, whenever the change in the function is proportional to the change in the variable, the function is linear. This is true, for example, in all cases of motion at a constant speed on either a straight or curved path ; the distance is then a linear function of the time. The following example will serve to illustrate the graphic method of solution. Example. At 7 a.m. a man starts to go up the 7-mile carriage road of Mt. Washington. At 9 o'clock he passes a party of ladies coming down. He reaches the top at 10 o'clock and, finding no view, he immedi- ately sets out on the return trip, which takes 1 f hrs. As he reaches the hotel from which he started he notices the party of ladies just arriving. At about what time did the ladies leave the top, assuming that the man kept up an approximately constant rate of speed on the way up and the ladles on the way down ? To solve the problem, we represent on a horizontal axis the time, mark- ing the hours 7, 8, 9, 10, 11, 12 and on the vertical axis the distances from the hotel at the foot of the t> m — I I ' I — I.I I I I I mountam. The graph' of the man going up the mountain is a straight line starting at O (at 7 a.m. he was at the hotel) and ending at a point A representing 10 o'clock and 7 miles from the hotel. Regarding the ladies, we know that the graph of their descent is also a straight line. At 9 o'clock they were the same distance from the hotel as the man. The point B on the line OA, corresponding to 9 o'clock, /> A N / \ >v / > iP / / \ / \ z / \ \ b£ 07 9 10 Fig. 35 11 12 houna Ill, § 58] LINEAR FUNCTIONS 79 is then one point of the ladies' graph. Another point is the point C at distance from the liotel at n:45. The line BC is then drawn and extended to Z>, representing 7 miles distance from the hotel. It is seen that the ladies started at about 7:30. How far was the man from the top when he met the ladies ? 58. Sum of Two or More Functions. Let mix -h h^ , m^x-\-h2 , '", m^x -f- 6^ be any k linear functions of x. The sum of these functions is (miX + 6]) -\-{m2X -[- 62)+ ••• H-(^fc» + &*) and this is equal to (mi+W2-f ••• +mj)x+{hi 4-62H h &jk)> which is again of the form mx + h. The result may be stated as follows : The sum of any number of linear functions of x is itself a linear function of x. Example. An empty tank is being filled by a faucet supplying 2 gallons of water per minute. After this faucet has been running 5 minutes a second faucet is turned on which supplies water at the rate of 3 gallons per minute. When the two faucets have been running to- gether for 6 minutes, an outlet is opened, but both faucets continue to Fig. 36 run. If the tank is empty at the end of 32 minutes, counted from the start, draw a graph representing the amount of water in the tank at any instant. Find approximately the rate of flow from the outlet, which may here be considered constant. 80 MATHEMATICAL ANALYSIS [III, § 58 We shall represent minutes on the horizontal scale and gallons of water in the tank on the vertical scale. The increase of water due to each faucet is at a constant rate, and the decrease when the outlet is opened is also at a constant rate. The amount of water in the tank due to each cause separately is, therefore, a linear function of the time, and their algebraic sum is also a linear function of the time. The first faucet begins at < = (when w, the amount of water in tank, is 0) to supply water at a uniform rate which would supply 40 gallons in 20 minutes. The amount of water in the tank due to the first faucet almie would then be represented at any instant by the straight line OA joining the points O (0, 0) to A (20, 40). The second faucet begins when t = 5 to supply water at the rate of 30 gallons in 10 minutes. If this second faucet were operating alone, the water supplied by it at a given instant would be represented by the straight line joining B (5, 0) to C (15, 30). In the actual problem from the instant i? = 5, the two faucets are running simultaneously. The sum of the two functions is then rep- resented by the line-segment DE, where Z> = (5, 10) and £' = (10, 20 + 15) = ( 10, 35). This line may be obtained graphically from the figure. When t = 11, a new factor enters, which reduces the amount of water in the tank to zero at i = 32. You may now finish the discussion. The required graph is the broken line ODHI. What would be the effect on the graph if one or both faucets were turned off at < = 20, the outlet remaining open ? EXERCISES 1. A man on horseback rides from a place J. to a place 5, 15 miles distant, in 2 hours. When he is 4 miles from A, he passes a lady walk- ing in the same direction. The man remains at B \ hour and then returns to A on foot. After walking 1 hour, he meets the lady on her way to B. If the man walks at the rate of 3 miles per hour, find the rate at which the lady is walking and at what time she left A. 2. A man starts at A to walk through B to a place C. At the same time a second man starts to walk from B to C. The first man reaches B in 1^ hours, while the second man has only walked f as far in this time. In how many hours will the first man overtake the second ? 3. Represent graphically on the same drawing the motion of the hour and the minute hand of a clock and use the drawing to determine ap- proximately at what time the two hands are in the same position. [Hint: The hands are together at twelve. Lay off the hours from 12 (or, 0) to 12 on the horizontal axis and the angles in degrees that either Ill, § 59] LINEAR FUNCTIONS 81 hand makes with the 12 o'clock position on the vertical axis. Each liand moves at a constant angular speed. The graph of the hour hand is then a straight line joining the points (0, 0), (12, 360). The minute hand goes from to 360 in 1 hour. The graph during the first hour is then a straight line joining (0, 0) to (1, 360). At 1 o'clock the graph begins at (1, 0) and goes to (2, 360) and so on.] 4. At what time between five and six o'clock are the hands of a watch together ? 5. At what time between two and three o'clock are the hands of a watch opposite to each other ? At right angles ? 6. At what time between four and five o'clock are the hands of a clock at right angles ? (Two solutions.) 7. A and B start to walk towards each other from two towns 15 miles apart. A walks at the rate of 3 miles per hour but rests one hour at the end of the first 6 miles. B walks 4 miles per hour but rests two hours at the end of the first 4 miles. In how many hours do the two men meet ? 8. Two men can do a certain piece of work in 12 and 15 days re- spectively. After the first man has worked 3 days alone, the two men finish the work. How long do they work together ? Ans. 5 days. 9. A messenger boy riding a bicycle at the rate of 9 miles per hour is sent to overtake a man on horseback riding 6 miles per hour. How long will it take the boy to overtake the man if the man had a start of 4 miles ? 59. Explicit and Implicit Functions. We have hitherto considered functions which were defined explicitly by an expression involving the variable. Thus the relation between y° Fahrenheit and x° Centigrade was expressed by the relation Now let us consider the equation 2x — Sy-{-7 = 0. This equation also defines a functional relation between two vari- 82 MATHEMATICAL ANALYSIS [III, § 60 ables. To every value of x corresponds a definite value of 2/, and, conversely, to every value of y corresponds a definite value of X. But, the equation does not express one of the variables explicitly as a function of the other. In fact the form of the equation gives no indication which of the variables is to be considered as the independent variable and which as the function. Such a relation is said to define a function implicitly. From such an implicit relation we can derive the expression of either variable as an explicit function of the other. Thus from 2iB— 32/-f7 = follows at once 2/ = l^+i and x=:f2/-f The first of these equations expresses y as an explicit function of X, and the second expresses x as an explicit function of y. 60. The General Equation i4x -h 5y + C = 0. Any linear relation between two variables x and y can be written in the form (1) Ax^-By-\-C^Qf. For example, the relation just discussed in the preceding arti- cle is obtained from this general relation by placing ^ = 2, 5 = — 3, (7=7. Equation (1) always defines ?/ as a linear function of x^ except when 5 = 0. In this case the term in- volving y drops out and the equation reduces to Ax + C = 0, and we cannot speak of 2/ as a function of x. But, if 5 :^ 0, we have By = —Ax—C^ or A C y = — ^ — J ^ B B' which is of the form y = mx -f h. Hence we conclude : Any equation of the form Ax + By -\- C =0 defines y as a lineal' function of x for all values of A, B, C except B = 0. Ill, § 61] LINEAR FUNCTIONS 83 61. The General Equation of the Straight Line. It follows from the result of the last section, that the locus of the equa- tion when interpreted geometrically in rectangular coordinates, is a straight line, except perhaps when B = 0, when the equa- tion takes the form Ax -\- C = 0. In this case, if A = also, the equation reduces to (7 = 0, and it completely disappears. If A is not zero, we may solve the equation for x and obtain, G . = --, or « = a constant. Now, the locus of a point whose abscissa is constant is a line parallel to the ?/-axis and at a distance equal to the constant from it. Thus the locus of a; = — 3 is a line parallel to the y-Sixis, and three units to the left of it. The case JB = is not then an exception, and we have the following theorem. Every equation of the form Ax ■\- By -\- C = 0, when repre- sented geometrically by means of rectangular coordinates, repre- sents a straight line. If B = 0, the line is parallel to the y-axis ; 1/^ = 0, the line is parallel to the x-axis; if (7=0, the line passes through the origin. Prove the last two statements of this theorem. We may also state the following theorems. Every straight line in the plane may he represented by an equa- tion of the form Ax -j- By -\- C = 0. The loci of Axi-By-{- C=0 and k (Ax -\- By -\- C) = (k =^ 0) a?'e identical. The proofs of these theorems are left as exercises. 84 MATHEMATICAL ANALYSIS [III, § 62 62. Analytic Geometry. We have thus far used the notion of coordinates to give a geometric interpretation to algebraic relations. It is possible to reverse the process and use the connection established between algebra and geometr}^, for the study of geometry. This method of studying geometry by algebraic means is called analytic geometry. In the following sections we proceed to develop certain analytic methods applicable to the straight line. The results are, in a large measure, merely a restatement from a different point of view of the results already obtained. 63. Straight Lines. We have already seen that the graphs of equations Ax -\- By -\- C = and y = mx + & (§ 52), when represented by means of rectangular coordinates, are straight lines. In § 60 we saw that the first of these equations could be put in the form of the second, provided 'B ^ 0. Thus when an equation of the form Ax -{- By -\- C = is solved for y, the coefficient of x is the slope, and the constant term is the y-intercept. The slope of the line connecting the two points Pi(iCi, 2/i), A(aJ2,2/2) is (§§51-53) We see geometrically that a line is determined when we know its slope and a point on the line. To determine the equation of this line, if {x^, y^) is the given point and m the given slope, we proceed as follows. Let {x, y) be any variable point on the line. Then, equating slopes, we have X — Xi that is, y-y,= m{x- X,) is the required equation. Ill, § 64] LINEAR FUNCTIONS 85 It is left as an exercise to prove that the equation of the straight line through the two given points (x^, y^), (ajg, 2/2) is if Xi =^ X2. 2/-yi = ^'_^' (^-^i)> 64. Parallel Lines. In Fig. 37 let (1) and (2) be two parallel lines with slopes mi and mg. Construct the positive segments PiQi and P2Q2 from the points P^ and P2 on lines (1) Fig. 37 and (2) respectively, and complete the right triangles P^QiRi and P2Q2R2' We then have m , = ^1^ and m2 = ^'^' If the lines are parallel, QiRi and Q2R2 are either both positive or both negative ; m^ and mg have then the same sign. They have the same magnitude since the triangles P^QiRi and P2Q2R2 are similar. Hence, If two lines are parallel, their slopes are equal, i.e. m^ = mj. Conversely, if the slopes of two lines are equal, the lines are parallel. The proof of this statement is left as an exercise. 86 MATHEMATICAL ANALYSIS [III, § 65 7 V { 9 A y \ \ X Fig. 38 vertical line B1R2 . 65. Perpendicular Lines. In Fig. 38 let (1) and (2) be two perpendicular lines with slopes mi and rriz and let the units on the two axes be equal. From the intersection P of the two lines construct the positive horizontal segment PQ of any convenient length. Through Q draw the We then have m, = ^Ma„dm, = «^ PQ PQ Therefore the slopes have opposite signs. Why ? Also from the right triangle R.R^P we have P^ = | QE^ \ • | QR2 1. There- fore * mim2 = — 1 and m, =- That is, if the units on the coordinate axes are equal , perpendicu- lar lines have slopes ivhich are negative reciprocals of each other. Conversely, if the slopes of two lines are negative reciprocals of each othery the lines are perpendicular, provided the units on the coordinate axes are equal. The proof of this statement is left as an exercise. Why is it necessary to assume the units equal ? 66. Illustrative Examples. Example 1. Find the equa- tion of the straight line through the point (4, 7) and having the slope — 2. * This proof presupposes that neither mj nor mg is zero, i.e. the lines are not parallel to the coordinate axes, and the result obtained does not apply to such lines. However, two lines parallel to the x- and y- axes have equa- tions of the form y = a constant and a; = a constant, respectively, and henco can be recognized at once. Ill, § 66] LINEAR FUNCTIONS 87 We have at once from § 63, ?/ — 7= — 2 (ic — 4) or 2 a; -f 2/ - 15 = 0. Example 2. Find the equation of the straight line through the points P^{2, 4), P^{- 5, 6). rru 1 6-4 2 The slope m = - — - — - = — 5—2 7 From § 63 the equation of the line is y ~ 4:= — ^ (x — 2) ov 7y + 2a;-32 = 0. Example 3. Express the temperature measured by y° Fah- renheit as a function of fl;° Centigrade. We know that when y = 32, x = 0: i.e. Pi(0, 32) is a point on the graph. In the same way we have PaC^^O, 212) a point of the graph. Therefore the equation of the line connecting these points is ' 212-32 ^ y-32 100 - a; - or 2/ = 9 a; + 32 (See § 55, Example 1.) Example 4. Find the equation of the straight line through the point (2,-5) and parallel to the line 2 y + 4 ic — 5 = 0. The slope of the given line is — 2 (§ 63). Therefore the equation of the required line is y-{-5= — 2(x — 2) or 2a; + 2/ + 1 = 0. Example 5. Find the equation of the straight line through the point (1, — 2) and perpendicular to the line 3x — y -\-2 = 0. The slope of the given line is 3. Therefore the slope of the required line is — ^ (§ 65). The equation of the required line is2/-f2 = — i(a; — l)ora;-f32/4-5 = 0. 88 MATHEMATICAL ANALYSIS [III, § 66 EXERCISES 1. What is the meaning of the constants m and h in the equation y = mx + b? 2. What is the effect on the line y = mx + & if 6 is changed while m remains fixed ? If m changes when b remains fixed ? 3. Describe the effect on the line y — yi = m(x — xi) if m changes while xu y\ remain fixed : also describe the effect if Xi, yi, vary while m remains fixed. 4. What is the equation of the line (a) whose slope is 3 and whose ^-intercept is 2 ; Ans. y = 3 x + 2. (6) whose slope is 4 and whose y-intercept is — 3 ; (c) whose slope is and whose ^/-intercept is — 1 ; (d) whose slope is and whose y-intercept is ? 5. Describe the positions of lines (c) and (d) in Ex. 4. 6. Define " y-intercept of a line." What is meant by the "x-inter- cept"? 7. For each of the following lines give x-intercept, y-intercept, and slope : (a) 2x-3?/ = 7. ^ns. I; -I; f. (c)2x-y + 5 = 0. (6) X + y - 2 =: 0. ((?) 4 X + 2/ = 0. 8. Is a straight line determined if we know its intercepts ? Try each of the equations 2 x — y = 4 and 2 x — y = 0. 9. Find the equation of the line joining the two points (2, 1) and (—3, 1); of the line joining the points (4, 2) and (4, —3). 10. Which of the following lines are parallel ? (a) 2 X - 2/ - 4 = 0. (c) 4 X - 2 ?/ - 1 = 0. (6) y + 2 X + 3 = 0. (d) 2 y + 4 X + 5 = 0. 11. Are the points (1, 5), (- 1, 1), (2, 6) on the line y = 2 x + 3 ? 12. What is the equation of the line which is parallel to y = 2 x + 3 and passes through the origin ? perpendicular to y = 2 x + 3 and passes through the origin ? 13. Determine k so that (a) the line 2x + 32/ + A; = shall pass through the point (0, 1) ; Ans. - 3. (b) the line 2x-f-3y4-* = shall have a y-intercept equal to 2 ; {c) the line 2x + 3y + ^ = shall have an x-intercept equal to 5. Ill, § 66] LINEAR FUNCTIONS 89 14. Find the equations of the sides of the triangle whose vertices are (3,4), (-1, 2), (-4, -5). Ans. x-2y + 5 = 0; 9x-ly + l=0- 7x-3i/+13 = 0. 15. Find the equations of the sides of the quadrilateral whose vertices are (-2,1), (3, -1), (- 2, 4), (1, 7). 16. What intercepts does the line through the points (2, — 7) and (4, — 5) make on the axes ? 17. Find the equation of the line which passes through the point (4, — 2) and whose slope is 6. 18. A line has the slope 2 and passes through the point (—1, 2). What are its intercepts ? 19. What is the equation of the line which passes through (—6, 5) if its y-intercept is — 3 ? Ans. 8a: + 5?/ + 16 = 0. 20. Write the equations of the lines which make the following inter- cepts on the X- and !/-axes. (a) 2 and - 4 ; (b) - 7 and - 3 ; (c) 4 and 5 ; (d) and 0. 21. If the X- and y-intercepts of a line are a and b, prove that the equa- tion of the line can be written iii the form a b [This equation is called the intercept form of the equation of a straight line.] 22. Solve Ex. 20 by using the result of Ex. 21. Does the formula hold in Ex. 20, (d) ? Explain. 23. Find the equation of the straight line through the point (4, — 6) parallel to the line 2x— y-{-7 = 0; through the same point, perpendicu- lar to the line 2x — y + i = 0. Ans. y = 2x— IS; 2y = — x — Q. 24. Prove that the lines Ax + By -\- C =0 and Ax -{- By + D = are parallel. State this theorem in words. 25. Prove that the lines Ax -\- By + (7 = and Bx—Ay-^D = are perpendicular. State this theorem in words. 26. Prove that the lines Ax -\- By + = and Mx -h Ny -\- P = Sive perpendicular if and only if AM -{■ BN = 0. 27. Show that the points (- 8, 0), (-4, - 4), (-4, 4), and (4, - 4) are the vertices of a trapezoid. 28. The Reaumur thermometer is graduated so that water freezes at 0° and boils at 80°. Find the equation of the line that represents the read- 90 MATHEMATICAL ANALYSIS [III, § 66 ing B of the Reaumur thermometer as a function of the corresponding reading C of the Centigrade thennometer. 29. A printer asks 75 cents to set the type for a notice and 3 cents per copy for printing. The total cost is what function of the number of copies printed ? Draw the graph of this function. 30. Express the value of a $ 1000 note at 6 % simple interest as a function of the time in years. Is this a linear function ? 31. A cistern is supplied by a pipe that supplies water at the rate of 30 gallons per hour. Assuming that the amount A of water in the cistern is connected with the time ^ by a linear relation, find this relation if ^ = 1000 when t = 10. What is A when t = 0? 32. In stretching a wire it is assumed that the elongation e is con- nected with the tension t by means of a linear relation* Find this rela- tion if i = 20 lb. when e = 0.1 in. and t = 60 lb. when e = 0.3 in. 67. Systems of Straight Lines. An equation of the first degree in x and y, and containing an arbitrary constant, repre- sents in general an infinite number of straight lines. For the equation will represent a straight line for each value of the constant. All the lines repre- sented by an equation of the first degree containing an arbitrary con- stant are said to form a system of lines. The arbitrary constant is called the parameter of the system. Thus the equa- tion yz= — 3x-{-b represents the system of straight lines with slope — 3. (See Fig. 39.) The equation y — 2 = 7n{x — l) represents the system of straight lines through the point (1, 2).* (See Fig. 40.) * It represents every line of this system except the one parallel to the y-axis. Why ? Fig. 39 1 c : i^.Z-t ^v i X - - ^:,: [V i y ' sA' *" J^S r-t -Q"~ ^ V :.^^^\n. i :: -5^ ii 1i-^ X ::: dZ ±i: :± Fig. 40 lU, § 68] LINEAR FUNCTIONS 91 68. Pencil of Lines. All the lines in a plane which pass through a given point are said to form a pencil of lines. The point is called the center of the pencil. 11 Ax + By ~\- C = 0, and A'x + B'y -\- C' = are any two lines of the pencil, then (3) (Ax -f % 4- 0)-h k(A'x + B'y + 0')= 0, where k is an arbitrary constant, represents a line of the pencil. This is true because the equation- (3) (a) is of the first degree in x and y and therefore represents a straight line ; (b) is satisfied by the coordinates of the point of intersec- tion of the two given lines. Why ? Example 1. Eind the equation of the line through the point (2, — 5) and parallel to4:X-{-2y-{-5 = 0. The system of lines parallel to 4:X -\-2y -\- 5 = is given by the equation y= — 2 x -\-7c. Since we want the particular line of the system that passes through the point (2, — 5), the equation must be satisfied by these coordinates. It follows that, — 5 = — 4-f-A:orA;= — 1. Therefore, y = —2x — 1 is the desired equation. Example 2. Find the equation of the line through the point (4, — 1) and perpendicular toSx-\-2y — 5 = 0. The system of lines perpendicular to 3x-\-2y — 5 = is given by the equation y = ^ x -\- k. Since we want the line of the system that passes through the point (4, — 1), we have k = — y . Therefore, the desired equation is y=2x_i_i or 2a; -31/ -11 = 0. Example 3. Find the equation of the line through the intersection of 2x-\-y — 4: = and x-{-y—l=0, and perpen- dicular to x-\-2y = S. Any other line through the intersection of the given lines is (4) (2x -{- y - 4:)+ k (x + y - 1) =0 92 MATHEMATICAL ANALYSIS [III, § 68 or x{2 + k)-\- y{l + k) + {- i - k) = 0. The slope of this line is — (2 + k)/{l + A:) and this must be equal to the negative reciprocal of the slope of the straight line X -f 2 2/ = 3. Therefore, -2^^=2 and k=-^. 1 + k 3 Substituting this value in equation (4) and simplifying, we have 2x — y — S = 0, the required equation. EXERCISES 1. Find the equation of the straight line through the point (1, 5) and parallel to2x-\-Sy— 9=0; perpendicular to2x + 32/ — 9 = 0. Alls. 2x +3?/-17 = 0; 3x-2y+ 7 =0. 2. Find the equations of the altitudes of the triangle whose vertices are (2, 8), (4, - 5), (3, - 2). 3. Find the equation of the straight line through the intersection of 10x4-5^ + 11 = and x +2y + 14 = which is perpendicular to x + 7y + l=0; parallel io S x - 7 y = I. 4. Find the equation of the straight line through the intersection of X -\-2y — 4: = and x — Sy-^1=0 which is perpendicular to Sx — 2y + 4 = 0; parallel to x — y = 0. 5. Find the equation of the straight line through the intersection of X -\-y - 1 = 0, x-3y+4 = and (a) through the point (1, 1) ; Ans. x + 6y — 6 = 0. (&) parallel to the line x + 2?/ — 9 = 0; (c) perpendicular to the line 4 x — 5 ?/ = ; ((7) through the intersection of3x + 4?/ — 8 = and x— 5j/ + 7=0. 6. Find the equation of the straight line which passes through the point (a) (0, 0) and is parallel to2x — y + 4 = 0; (6) (1, 2) and is perpendicular to3x— 2^/ — 1 =0; (c) (— 1, 2) and is parallel to x — y — i=0. 7. Find the equation of the line which passes through the inter- section of X — 2^ + 2 = and x + y = and through the intersection of x + y+2 = 0, x-?/ = 0. Ill, § 69] LINEAR FUNCTIONS 93 8. Find the equation of the straight line through the intersection of x-2y -{-7 = and 2x-y-i-S = and (a) parallel to the aj-axis ; (6) parallel to the «/-axis. 9. Find the equation of the straight line which passes through the intersection ofSx— ^ + 2 = and x + y = 6 and which (rt) passes through the origin ; (b) is parallel tox — 4?/ + 3=0; (c) is perpendicular to 3aj — 2^ + 4 = 0. 69. Intersection of Two Lines. Simultaneous Equations. We have just seen that linear equations in one or two vari- ables are represented in rectangular coordinates by straight lines. We now wish to determine the coordinates of the point of intersection of two lines whose equations are given. That is, algebraically, we wish to find a set of values for x and y which satisfy both equations. Example 1. Solve the equations (6) (6) Sx — 4:y = 7. x-\-2y=9. m Multiplying equation (6) by 2 and add- ing the result to equation (5), we obtain 5 a; = 25, or X = 5. Likewise multiplying equation (6 ) by 3 pj^ ^i and subtracting the result from equation (5), we have — 10 y = — 20, or y = 2. The set of values x = 5, y = 2 is seen to satisfy both equations and is called the solution of the given equations. If we plot lines (5) and (6) (Fig. 41), we see from their graph that the coordinates of their point of intersection are (5, 2). Therefore, a method of solving two linear equations in one or two variables is to plot the lines represented by each equation, and then deter- mine from the graph the coordinates of the point of intersection. The algebraic method of first eliminating one variable and then the other has the advantage over the geometric method in that it is always accurate. Instead of eliminating twice, the value found for either variable can be substituted in either equation, and the value of the second variable de- termined. 94 MATHEMATICAL ANALYSIS [III, § 69 Example 2. Solve the equations (7) (8) T ~ ~ ~ n - C\ ^^ ^ ^A ^ •i % ^ ^' ^ - ?- gf* ^ ^ y ^ n Y" ■" _ _ _ x-2y = S. X — 2 ?/ = — 5. Subtracting the second equation from the first, we obtain = 8. Tliat is, there are no values of x and y satisfying both equations. Such equations are said to be inconsistent or incompatible. We see that lines (7) and (8) have the same slope, but different y-intercepts, and therefore are parallel lines. Example 3. Solve the equations Fig. 42 (9) (10) x-y = 2. 2x 2y=4. Multiplying the first equation by 2 and subtracting the second from it, we have 0=0. If equation (10) be divided by 2, equations (9) and (10) are seen to represent the same relation between x and y, and are not therefore sufficient to determine x and y. We can assign to either vari- able an arbitrary value and then find the corresponding value for the other variable. The equations can, therefore, be said to have an infinite number of solutions. Such equations are called dependent. The graphs of these equations are coincident lines. Let us now consider the general equations (11) aiic + bill = Ci , (12) a-^x-^ b^ = C2, where none of the constants are zero. Eliminating y, we obtain {a^bz — a2&i) x = C162 — Cobi . Eliminating x, we ob- tain {oibo — aib^) y = aiC2 — a^c^ . Now if 0162 — 02^1 =^ ^) we have C261 x^"^^^^ aiC2 — «2Ci we cannot aib2 — 0.2^1 ^1^2 ~ ^2^1 If, however, aib^ — aa^i = 0, i.e. a.^ja^ = &2^i) solve for x and y. Denoting the common value of these quo- tients by A:, we have ag = fcai , 62 = ^^1 • Then equations (11) and (12) become a^x + b^y = Ci , and "ka^x -f- Icb^y = Co . Ill, §69] LINEAR FUNCTIONS 95 We must now distinguish two cases according as C2 = kci or C2 ^fc kci . In the former case, by dividing out k, we see that the equations are dependent and have an infinite number of solu- tions. In the latter case they are inconsistent, and thus are not satisfied simultaneously by any values of x and y. Discuss the cases that arise if some of the constants are zero. EXERCISES Find, when they exist, the coordinates of the points of intersection of the following lines. Check your answer from a graph. 1. 4a; + 2?/=:9. 3. x + 2y = 3. 5. a: + 4?/ = l. 2x-5y = 0. 2x+4y = 6. 2x + 8?/ = 2. 2. 3x + 4?/ = 12. 4. x-2y = 7. Z.x-^y = 1. X — y = b. 2x — iy = b, — x -\-2y = 3. In the following exercises are the lines concurrent ? If so, what point have they in common ? 7.x + 2?/ = 3. 8.x-y = -l. 9. x+2y = 5. 10.x-2y = 3. x — y = 0. 2x-{-y = S. 6x— y = 3. 5 x — y = 2. 5x— ?/=4. Sx-2y = l. 2x + y=4. 2x + 3?/ = l. In the following exercises, find k so that the lines shall be concurrent. ll.x+y = 2. 12.2x-y-0. 13. 3 x - y = 4. 2x—y = l. x + 3y = 7. x + y = 0. 4x + y = k. Ans. b. x -^ ky = b. b x — 2 y = k. 14. The sides of a triangle have for their equations 2 x + y = 5, a;— ?/ = 10, — 2x + ?/ = 6. What are the coordinates of the vertices of this triangle ? What are the equations of the altitudes ? 15. Find the equation of the straight line through (2, 1), (—1, 2), using the equation Ax + By + C = 0. [Hint : Solve for A/C and B/C] 16. Find the equation of the straight line through (4, 7) and having the slope 3, using the equation Ax + By -\- C = 0. 17. It has been shown experimentally, that the length I of a wire in feet under a tension of p pounds, is I = a + bp, where a and b are constants. Find a and bii 1= 190 when p = 270, and that I =190.2, when p = 450. 18. The readings T and 8 of two gas meters are connected by the equation T = a -}- bS. Determine a and b when we know that 8 = 10, when T = 300, and S = 100, when T= 420. 96 MATHEMATICAL ANALYSIS [III, § 69 19. The pull in pounds to lift a load I in pounds with a pulley is given by the relation p =: al + b, where a and h are constants. Find a and b when it is known that a pull of 8 pounds lifts a load of 40 pounds, while it takes a pull of 2 pounds to hold the rope on when no weight is attached. 70. Equations Containing More than Two Unknowns. It is easy to see that the methods employed in § 69 for solving a system of two simultaneous equations, each containing two unknown quantities, may also be employed for solving a system of three or more equations, involving as many unknown quantities as there are independent equations. Example. Solve the equations (13) 7x-\-Sy-2z = 16. (14) 5x-y-\- 6z = Sl. (15) 2x + 6y + 3z = S9. Adding three times (14) to (13) gives (16) 22 X + 13 ;s = 109. Adding five times (14) to (16) gives (17) 27 a; + 28^= 194. Solving equations (16) and (17) by the methods of § 69, we have a; = 2, z = 5. Substituting these values in (13), we obtain y = i. It is readily seen that x =2^ y = 4, z = b satisfies equations (13), (14), (15). The cases in which three simultaneous equations in three unknowns have no solution, or an infinite number of solutions, will be discussed in Chapter XXI. EXERCISES Solve the following simultaneous equations : l.->2x-\-4y-{-z = 12. 2. X + y + z = IS. 3. 2x -Sy - z = 2. ^^x-\-y-z = S. x-2y-{.iz = l0. 6x-i-2y + z = -8 x -\- y + z = 7. 3x + y~Sz=5. x—2y — z = 2. 4. a; + 8 2/ - 4 = 9. 5. w + a- 4- ?/ = 15. 6. a: + y = 4. 3x+3?/ — = 6. x + y -{■ z = 19,. 2x-\- z = 4. bx + 2y-2z = l. wj + ?/ + 0=17. y-2 = 3. 10 + X + = 16. Ill, §70] LINEAR FUNCTIONS 97 7. If A and B can do a piece of work in 10 days, and A and C in 8 days, and B and C in 12 days, how long will it take each to do the work alone ? 8. Three towns A, B, and C are situated at the vertices of a triangle. The distance from A to B via C is 76 miles ; from A to C via B 79 miles; from B to C via A 81 miles. Find the distance from A to B, from B to C, from C to A. 9. In a triangular track meet the following was the final score : SCOKE First Place Second Place Third Place Total College A . . . College B . . . College C . . . 5 2 2 3 4 2 3 1 6 37 23 22 How many points did each place count ? 10. Two passengers traveling from town A to town B have 500 pounds of baggage. The first pays $ 1.75 for excess above weight allowed, the second $1.25. If the baggage belonged to the last passenger, he would have to pay $ 4 excess. How much baggage is allowed to a single passenger ? 11. A crew can row 4 miles downstream and back again in 1| hours, or 6 miles downstream and half way back in the same time. What is the rate of rowing in still water, and what is the rate of the current ? Ans. 6 miles per hour ; 2 miles per hour. 12. Two trains are scheduled to leave two towns A and B, m miles apart, at the same time, and to meet in h hours. The train leaving A was k hours late in starting, so the trains met n hours later than the scheduled time. What is the rate at which each train runs ? 13. Two men are running at uniform rates on a circular track 150 feet in circumference. When they run in opposite directions, they meet every 5 seconds. When they run in the same direction, they are abreast every 25 seconds. What are their rates ? 14. Find a, &, c, so that y = a -\- bx -\- cx^ shall be satisfied by (2, 1), (1,0), (3,-6). 6a;2_ic-3 a 16. Find a, 6, c, so that x^ X + 1 X, CHAPTER IV THE QUADRATIC FUNCTION I. GRAPHS OF QUADRATIC FUNCTIONS 71. The General Quadratic Polynomial ax^ -h bx -{-c. Having considered in some detail the linear function mx + b and its geomietric interpretation, we now turn our attention to a similar study of the quadratic function, i.e. sl function ex- pressed by a polynomial of the second degree in one variable. Such polynomials are, for example, a;^ -f- 1, 100 -f- 50 ^ — 16.1 f^j etc. The general form of such a polynomial is ax^ -\- hx^ -f c, where a,h, c are constants and a^O. Such functions abound in practice. Thus, if a projectile be shot vertically upward from the top of a tower 100 ft. high, with an initial velocity of 50 ft. per second, the distance s (in feet) from the ground at the end of t seconds, is given approximately by the poly- nomial s = 100-f 50i-16.1f2. The general formula for the distance s from the ground at the end of t seconds of a projectile shot vertically upward is (approximately) where Sq is the distance from the ground when t = 0, Vq is the initial velocity, and g is the so-called " gravitational constant," which varies slightly from place to place but is approximately equal to 32.2 when the distance s is measured in feet and the time is measured in seconds. IV, § 72] QUADRATIC FUNCTIONS 99 72. The Function x^. We consider first the simplest of all quadratic functions, viz. the function y = x^. A brief tabular representation of this function is as follows : X 1 2 3 4 -1 -2 -3 -4 y 1 4 9 16 1 4 9 16 If we plot these points, we obtain Fig. 43, in which we notice that the points seem to be arranged according to some regular law. We may insert additional points by calculating values ~ Y * 1 ~\ ^///A t '/, V' z w, wM X V Fig. 43 of y for values of x between those already used. Thus for X = 1.5, y = 2.25 and a; = — 1.5 y = 2.25. These points are also marked on the figure. In general we see that for X = a and also for a; = — a, we have y = a^. . Geometrically this means that the graph is symmetrical with respect to the 2/-axis, i.e. if the part of the graph on the right of the y-axis is turned about the ?/-axis until it falls in the original plane, it will coincide with the part on the left of the y-Sixis. More- over, since x"^ is positive (or zero) for all real values of x, no part of the graph will be below the a;-axis. 100 MATHEMATICAL ANALYSIS [IV, § 72 Keeping these facts in mind we shall make a more detailed study of this function and its graph, by considering values of X which are closer together. We shall confine ourselves to values of x between x = and x = 2. The corresponding values of y, for all values in this range at intervals of 0.1 of a unit, are given in the following table : X y X y X y X y 0.1 0.01 0.6 0.36 1.1 1.21 1.6 2.56 0.2 0.04 0.7 0.49 1.2 1.44 1.7 2.89 0.3 0.09 0.8 0.64 1.3 1.69 1.8 3.24 0.4 0.10 0.9 0.81 1.4 1.96 1.9 3.61 0.5 0.25 1.0 1.00 1.5 2.25 2.0 4.00 We cannot, with any accuracy, insert in Fig. 43 the corre- sponding points of the graph. We therefore adopt a pro- cedure analogous to the use of a magnifying glass, in order to separate the points. This we have done in Fig. 44 by choos- ing the unit on each axis 10 times as large as in Fig. 43. We then see that there is no difficulty in plotting all the points given in the above table. Let us study more carefully the immediate neighborhood of some point on the graph, for example, P(l, 1). We shall magnify the shaded area in Fig. 44 in the ratio 10 : 1 and make use of the following table : X y X y X y X y X y 0.90 .8100 0.95 .9025 1.00 1.0000 1.05 1.1025 1.10 1.2100 0.91 .8281 0.96 .9216 1.01 1.0201 1.06 1.1236 0.92 .8464 0.97 .9409 1.02 1.0404 1.07 1.1449 0.93 .8649 0.98 .9604 1.03 1.0609 1.08 1.1664 0.94 .8836 0.99 .9801 1.04 1.0816 1.09 1 1.1881 IV, § 72] QUADRATIC FUNCTION-S 101 It will be noted that the points of the graph now lie almost on a straight line (Fig. 45). We have drawn a straight line through P for the purpose of comparison. If we should desire a more detailed representation in the neighborhood of the point P, we should calculate the values of y for values of x between x — .99 and x = 1.01 and draw anew a small portion w- 1.2 1.1 1.0 0.9 i 1 2 Fig. 44 .90 1.0 Fig. 45 1.1 of the figure about P under a tenfold increase of the unit. We would then find that the points would hardly be distin- guishable from the points on a straight line. Similar conclusions might be reached near any other point on the graph. It is of course impossible to prove this for each separate point by separate calculations. To prove the fact generally we proceed as follows. Let x^ be any particular value of the variable x and y^ the corresponding value of the function y ; then y^ = x^^. Now suppose that the value x is increased or decreased by a certain amount, which we shall call Ax (a decrease means that Ax is negative). The new value of X is then Xi -f Ax and the corresponding value of the 102 MATHEMATICAL ANALYSIS [IV, § 72 function is {xi + Aa;)2, This new value of the function differs from the original value of the function, yi, by a certain amount which we shall call Ay. We then have 2/1 + A?/ = (xi -h Axy = Xi_^4-2xiAx + Ax"^', but 2/1 = a^i Therefore, by subtraction, or (1) Ay = 2 cciAic + Ax'^ Ay = (2xi-\- Ax) Ax. Since formula (1) is true for every value of Xi , it follows that Ay approaches zero when Ax approaches zero. This means that in the neighborhood of the point {x^ , y^) we can find new Fig. 46 points on the graph whose x and y differ from those of the given point by as little as we please. This simply means that the set of all points of the graph oi y = x^ form a set of points with no gaps between them ; they form what we may call a continuous line or curve.* * A function is said to be continuous for a value x = Xi, if when Aa; ap- proaches the corresponding Ay also approaches 0. See footnote on p. 19. IV, § 73] QUADRATIC FUNCTIONS 103 Further, equation (1) gives the relation, ^ = 2 iCi + Ax (if ^x 4- 0). Aa; From the graph (Fig. 46) we clearly see that this change ratio is the slope of the line joining the points P^ix^ , y^) and P2(i>^i + Aa;, 2/1+ A?/).* If the latter point approaches the former along the curve, i.e. if we let Ace become numerically smaller and smaller, then the change ratio A?/ / Ao? will differ less and less from 2xy. Indeed, we may choose Aic sufficiently small (without making it zero) so that A?/ / Ax- will differ from 2 x^ by less than any previously assigned amount. Geometrically this means that in the immediate neighbor- hood of the point Pi on the graph oi y — a?^, the points of the graph lie very near to the straight line through Pi whose slope is 2 x^. From a somewhat different point of view, we can let the secant joining the points Pi {x-^ , y^) and Pg {xi-\-Ax, yi-\-Ay) on the graph rotate about Pi in such a way that Aic, and there- fore Ay, become smaller and smaller and the secant approaches a definite position. through Pi whose direction has the slope 2 Xi. This line is by definition tangent to the graph at Pj , or the graph is tangent to the line at Pi ; the point Pi is called the point of contact. Combining the above results we have : The graph of the function y = x"^ is a continuous curve, above the X-axis, symmetrical with respect to the y-axis, and passing through the origin. At any point Pi {xi , i/i) on the curve, the straight line ivith slope 2 Xi passing through this point is tangent to the curve. 73. Further Observations regarding the Function y = x\ The preceding result tells us that when x= 1, the slope of the tangent is 2. Reference to Fig. 45 will verify this result for * This follows also directly from the formula m = {yz— 2/i)/(x« — ^i)- 104 MATHEMATICAL ANALYSIS [IV; § 73 the straight line there drawn, since this line has the slope 2. In Fig. 47 we have reproduced Fig. 43 except that we have replaced the several points plotted in the earlier figure by a continuous curve and have drawn the tangent at the point P(l, 1). Knowing that the slope of the tangent is 2, we can easily construct the tangent. Starting from P we lay off any convenient distance PM to the right and then lay off double this distance MQ upward. The line PQ is then the required tangent. A similar process leads to the construction of the tangent at any other point of the curve. From the fact that the slope of the tangent at any point on the curve whose abscissa is Xi is 2 a^i, we see that as Xi increases numerically the slope increases numerically, that is, the curve becomes steeper and steeper the farther we go from the origin. Also the slope is positive when x^ is positive and negative when Xi is negative. This means that going from left to right the curve slopes downward at the left of the origin, and upward at the right of the origin. When a; = 0, the slope is zero, that is to say, the tangent is parallel to the a;-axis (here it coincides with the aj-axis). IV, § 73] QUADRATIC FUNCTIONS 105 Hitherto in our drawings we have chosen the unit on the t/-axis to be equal to that on the cc-axis. This renders it im- possible to draw the graph of the function y = x"^ for large values of a;, without making it of unwieldy size. However t I t 7 . _ - t 7 I } ^ ^ ^ ^ t Z r / ^z~ - z^i- // en J -U t ^ ^ "I ::^^: : : : A-- -«-- i' X ^ J ir :: ~A~r- : : : ^<^ :- = r'V.. _ . 23456789 10 Fig. 48 nothing prevents us from choosing the unit on the t/-axis smaller than that on the ic-axis, and in Fig. 48 we have chosen it one tenth as large. A tabular representation is as follows : X ±1 ±2 ±3 ±4 ±5 ±6 ±7 ±8 ±9 ±10 y 1 4 9 16 25 36 49 64 81 100 In this case the slopes of the tangents are, respectively, ±2, ±4, ±6, ±8, ±10, ±12, ±14, ±16, ±18, ±20. We have drawn the tangent at the point for which x = 5, and have drawn the graph only for positive values of x. Example. Find the equation of the tangent to the graph of 2/ = a;2 at the point (3, 9). The slope of the tangent at the point (x^ , 2/1) is 2 a?] . There- fore at (3, 9) the slope is 6. The equation of the tangent is, therefore, y — 9 = 6{x — S) oy y = 6x — 9. 106 MATHEMATICAL ANALYSIS [IV, § 73 EXERCISES 1. Discuss the functions y = — x^ ; y = 2x'^ \ y =— 2x^. 2. Construct for the point (2, 4) of the function y = x^ n. figure anal- ogous to Fig. 46. (Use a table of squares.) 3. Use the adjoining figure to give a geometric interpretation of the X Ax equation Ay = 2 xiAx i- Ax"^. The function y —x'^ is Ax ^^^® interpreted as the area of the square whose side is X. f////////M % % Ax 4. If in the function y = x^ we take x = 3, how small must Ax be taken in order that Ay shall be numerically less than 0.01 ? if we take x = 15 ? Is the difference between these two results to be expected in view of the nature of the graph ? 5. Draw the tangents to the curve 2/=x2 at the points for which x=0, ± i, ± 1, ± -I, ± 2. 6. If X is the radius of a circle and y is its area, prove that the change ratio Ay / Ax approaches the length of the circle as Ax approaches zero. 7. Find the equations of the tangents to the curve ?/ = x^ at the following points : (1, 1); (2,4); (-1, 1); (-2, 4). Construct the tangents at these points. 8. The line perpendicular to the tangent at the point of contact is called the normal to the curve at this point. Find the equations of the normals to y = x^ at the points (1, 1) ; (2, 4) ; (— 1, 1) ; (— 2, 4). Con- struct each normal making use of its slope. Ans. For the point (1,1): x + 2y-S =0.* 9. Find the slope of the tangent to y = 3 x^ at the point whose abscissa is xi. What is the value of this slope at the point (1, 3) ? 10. Find the equations of the tangent and the normal (see Ex. 8) to y = 3x2 at the points (3, 27) ; (- 2, 12). 11. Find the points where the slope of the curve y = x^ has the values - 1 ; 2 ; 10. 12. 1 cu. ft. of water weighs 66.4 lb. What must be the diameter x of a cylindrical can such that 1 in. of water contained in it will weigh y oz. ? Plot the graph and find x when y = 50. Find y when x = 8. * Assuming the units on the axis to be equal. IV, § 75] QUADRATIC FUNCTIONS 107 74. The General Quadratic Function j/ = ax^ + 6x + c. We may now dispose of the general case. Let y = ax^ -\- bx + c {a =^ 0) be any quadratic function (in the case y = x^, a was 1, while b and c were 0). Let x increase from the value Xi to the value Xi -f Ax, and suppose that this change in the value of x changes the value of the function from 2/1 to yi -f Ay. We desire to calculate the value of Ay and of the change ratio Ay /Ax. We have 2/1 -f Ay = a(x^ + Axy-\- b(xi + Ax) + c, and 2/1 = a^i^ + b^i + c. Subtracting, we obtain (2) A?/ = (2axi-\-b -\- a Ax) Ax, and (3) ^ = 2aXi-\-b-{-aAx (ifAa;^0). Aa; Equation (2) shows that Ay can be made numerically as small as we please, by choosing Ax near enough to 0. Hence we may say: Every function of the form y = ax"^ -\-bx+ c is continuous. Equation (3) shows that the change ratio Ay /Ax approaches as a limit the value 2axi + bsiS Ax approaches 0. Hence we may say: The slope of the tangent to the curve y = ax^ -\- bx -{- c at the point whose abscissa is Xi is equal to 2aXi 4- b. 75. General Properties of the Function ax^ + 6x 4- c. The discussion in the preceding section and the exercises have furnished us with some information regarding some special functions of the form aa^ -{-bx -\- c. 108 MATHEMATICAL ANALYSIS [IV, § 75 It will now be shown that whenever the term in ar* is posi- tive {i.e. a is positive) the graph of the function is an inverted a >o a < o Fig. 49 Fig. 50 arch as in Fig. 49 and that whenever the term in x^ is negative {i.e. a is negative) the graph is an arch like the one in Fig. 50. To prove this we need only consider the slope of the tangent to the curve as the point of contact moves along the curve. We have just seen that the slope of the tangent is given by the formula m — 2axi + & at the point whose abscissa is Xi . There is just one point on the curve for which this slope is zero, viz. the point whose abscissa is ^ a Now let us write the slope m of the tangent in the form The number in the parenthesis, i.e., Xi -f 5/(2 a), is positive when Xi> — 6/(2 a) and negative when Xi < — 6/(2 a). Geomet- rically this means that this parenthesis represents a positive number for points to the right of the straight line x= —6/(2 a) and a negative number for points to the left of this straight line. Case 1 : a > 0. If a is positive, the slope m is positive for points to the right of the line x = — 6/(2 a) and negative for points to the left of this line. In other words, for all points of the graph to the left of the line x = — b/(2 a) the tangent slopes downward (as we go from left to right) and for all points to the right of this line the IV, § 76] QUADRATIC FUNCTIONS 109 hi/>o Fig. 51 tangent slopes upward. The point for which a: = — 6/(2 a) has its tangent parallel to the aj-axis. This point is called the minimum point of the graph (Fig. 51). Case 2 : a < 0. Suppose on the other hand that a is negative. The slope m is then negative when Xi + &/(2 a) is positive and positive when x^ + 6/(2 a) is negative. The slope is therefore positive when Xi< — 5/(2 a) and negative when Xi> — b/(2 a). At the single point for which X — — (6/2 a) the tangent is parallel to the aj-axis. This point is called the maximum point of the graph (Fig. 52). When x = — 6/(2 a) the function y = ax^ -\- bx -\- c has a minimum value if a > and a maximum value if a < 0. The curve represented by the function y = ax^ 4- 6a; + c is symmetrical ivith respect to the line x = —b/(2 a). The proof is left as an exercise. Hint. Show that the points which have abscissas — 6/(2 a) + h and — 6/(2 a) — h have the same ordinate. Fia. 52 76. Definitions. of the form The curve represented by an equation y = ax"^ -{- bx -{- c is called a parabola. The lowest (or highest) point on this curve, i.e. the point for . which x = — 6/(2 a), is called the vertex. The straight line through the vertex and per- pendicular to the' tangent at the vertex is called the axis of the curve. The parabola is symmetrical with respect to its axis. 110 MATHEMATICAL ANALYSIS [IV, § 77 77. to draw the Graph of a Parabola y=ax'^-\~bx-^c. The preceding discussion enables us to draw the graph of a quadratic function without plotting many points. Example 1. Sketch the graph of y = 2 x^ — 6 x -\- 5. The slope of the tangent at {x^, y^ is, by § 74:^ m = 4:Xi — 6. The vertex of the curve is the point for which 4 ajj — 6 = 0, i.e. the point for which x^ = 3/2 ; the corresponding value of y is 1/2 and the vertex is therefore the point (3/2, 1/2). This point is the minimum point of the curve. We plot this vertex V, draw the horizontal tangent at this point and the vertical axis. We desire a few more points and their tangents on each side of the axis and then we can draw the curve. For ex- ample, we have Fig. 53 X y m 1 1 -2 2 1 2 5 -6 Example 2. Sketch the graph of 2/ = — aJ" + 4 a; -f- 5. The slope of the tangent at {xi , 2/i) is m = — 2 a^i -f 4. The vertex of the curve is at the point for which — 2 a^i + 4 = 0, i.e. for which Xi = 2. The corresponding value of 2/i is 9. Therefore the vertex, which is the maximum point of the graph, is at (2, .9). The graph is given in Fig. 54. 2 4 \ 6 Fig. 54 IV, § 77] QUADRATIC FUNCTIONS 111 EXERCISES 1. Tell which of the following functions have a maximum and which have a minimum value. Find this value in each case and the correspond- ing value of X. (a) 2x^ + Sx-9. Ans. Minimum value : — 17, when x =— 2. (&) 3 x2 -f 8 X - 6. (c) -6x^-\-l0x- 12. ((?) 3 a;2 + 6 X - 7. (e) -x^ + 1. 2. Find the coordinates of the vertex and the equation of the axis of each of the following parabolas. Sketch the curves. (a) 2/ = 2 ic2 + 5 X + 3. (6) 2^ = 3 x2 + 9 X - 6. (c) y = - 5 x2 + 10 X - 12. Ans. F= (1, — 7) ; axis, x=l. (d) ?/ = 3 x2 + 6 X - 7. (e) y=-x2+l. 3. The area of a certain rectangle in terms of the length of its side X is ^ = X (100 — 2 x). Find x so that this area shall be a maximum. 4. A point moves on a straight line so that its distance s from a fixed point O on the line at any time t is given by one of the equations below. Draw the (s, t) graph and in each case show that the variable point reaches, on one side of O, a maximum absolute distance from 0. Find this maximum distance. Does this maximum absolute distance correspond to a maximum or a minimum value of s ? (a) s = f2_4« + 3. * (6) s = 2fi-St + 10. (c) s = 3 + 6 « - 4 «2. 5. Find the equations of the tangent and the normal* to the curve y = x2 — 3 X + 1 at the point (1, - 1). Ans. y --x; y = x-2. 6. Find the equations of the tangent and the normal * to the curve y = — 2 x2 + 3 X - 1 at the point (1, 0) . 7. Find the equations of the tangent and the normal* to the curve ?/ = — 2 x2 + 4 X — 1 at the maximum point. Ans. y = I ; x = 1. 8. Find the equations of the tangent and the normal* to the curve y = 3x2 — 6x+lat its vertex. * See Ex. 8, p. 108. 112 MATHEMATICAL ANALYSIS [IV, § 78 78. The Graph of y - k = a{x- Kf. The face that the graphs of functions of the form y = ax- -f- 6a; 4- c, all have the same general shajje but are differently located with respect to the coordinate axes suggests that many of these graphs may consist of curves, which might be brought into coin- cidence by a suitable motion in the plane. That this is indeed the case results from the follow- ing considerations, which lead to a general principle of far-reach- ing importance. Suppose the graph of the equa- tion y = ax^ is moved parallel to itself through a distance and direction which carries the point to the point Q {h, k). What will be the equation between the x and y of any point P on the curve in its new position, the axes of coordinates remaining in their original positon ? This question is readily answered. Let P' be the position of P before it was moved. The equation y — ax^ then tells us that M^P — a • OM'^ for every position of P' on the curve in its old position. After the motion, the directed segments OM' and 3PP' become respec- tively the (Erected segments QR and EP. Hence, for every point P on the curve in its new position we have (4) RP=a'QR\ If the coordinates of P are (x, y) we have x = Oil/, y = MP and QR = x-h, RP=:y-k. Therefore, by (4), the curve in its new position is the graph of the equation (5) y — k = a ' (x — hy. While we have applied these considerations to the function IV, § 79] QUADRATIC FUNCTIONS 113 y = ax^, the reasoning is general ; consequently we may formu- late the following principle : GEJfEEAL Principle. If in any equation between x and y ice replace x by x — h and y by y — Jc, the graph of the neiu equation is obtained from the graph of the origiyial equation by moving the latter graph parallel to itself in such a ivay that the point moves to the point (Ji, k). We shall have occasion to apply this principle often in the future. 79. Transformation by Completing the Square. At present we may use the principle just stated to prove that the parabolas y = ax"^ -{-bx-^- c and y = ax^ are congruent curves. This follows at once from the preceding general principle, if we prove that the equation (6) y = ax^ -\-bx-\-c can be written in the form (7) y-Jc= a{x - hy. To do this we write (6) as follows : 2, = a(^ + ^+ )+o, and then complete the square on the terms in the parentheses by adding the term If' / (4 a^). In order to leave the value of y unchanged we must also subtract a y. b^ / (4a2) = b^ /{4^a) from the expression. This gives rrr,^ f . ,bx , b^ \ , ¥ (7') y = a\x'^ + —-' ' • ~ 4^^;"^' 4a' or = a[x-{- =) ^ ' 4a V 2 This is of the form (7) for the values 114 MATHEMATICAL ANALYSIS [IV, § 79 2 a 4tt The operation just performed is called the transformation by completing the square. It is found serviceable in a variety of situations. It may be used to advantage in connection with numerical examples. Example. Discuss the graph of y = — 2 x^ + S x— 9. We first write y=-2(a:2-4x+ )-9; and tiien ?/=-2(:c2-4.x + 4)-9 + 8 or 2/ + 1=- 2 (a; -2)2. The graph is then obtained from the graph ot y =—2x^ by moving the latter parallel to itself so that its vertex moves to the point (2, —1). EXERCISES 1. By reducing to the form y — k = a {x — h)^, discuss the graphs of each of the following functions. (a) y = 2x'^+12x-\-2. (d) y = 2x'^ - 1 x + S. (6) 1/ =4x2 + 6a:- 9. (e) y =- Ax"- + 1 x +2. (c) ?/ = - 3 x2 + 9 X + 10. (/) y = - 3 x2 - 8 X + 10. 2. Show that the equation of the straight line y — yi = m (x — xi) may be derived from the equation y = mx by the general principle of § 78. 3. The results of § 79 furnish a proof of the fact previously derived, that the vertex of the parabola y = ax2 + 6x + c is at the point for which x=— &/(2a). Explain. 4. Equation (7') proves that if a > 0, the value x = — 6 / (2 a) gives the minimum value to y ; also that if a<0, the value x=— 6/(2a) gives the maximum value to y. Explain without using the graph. Write the following equations in the form a (x — hy + 6 (y — ^')2 = c, where a, b, c, h, and k are constants. 6. x2-4x + 2?/2-8 2/ = 2. 8. x^ + y^-iy = 2. Ans. (x-2)2 + 2(2/-2)2 = 14. 9. a;2-8x + y2 = o. '6. -2x2 + 4x+?/2_4y_3_0. 10. 3x2-4x-y2 +2 =0. 7. 4 x2 - 4 X + 2 y2 _ 3 y ^ 1 _ 0. IV, § 80] QUADRATIC FUNCTIONS 115 II. APPLICATIONS OF QUADRATIC FUNCTIONS 80. Maxima and Minima. We have seen that a quadratic function ax^ -\-bx + c has either a maximum or a minimum value according as a is negative or positive. Numerous appli- cations involve the problem of finding this maximum or mini- mum value and the corresponding value of a;, as the following examples show. Example 1. A rectangular piece of land is to be fenced in and a straight wall already built is available for one side of the rectangle. What should be the dimensions of the rectangle in order that a given amount of fencing will inclose the greatest area ? Before beginning the solution proper we should note carefully the sig- nificance of the problem. The length of the fence being given, we may use it to inclose rectangles of a variety of shapes, as indicated by the dotted lines in Fig. 56. Some rectangle whose shape is between those indicated will inclose the maximum area. |.^,,^,,,,,,,,^^^^^^^^^ 1 X I — I Fig. 50 To determine this shape is our problem. To do this, it is necessary to express the | ' y j area (the quantity we wish a maximum) as a function of one variable. Solution : Let the dimensions of the rectangle be x and y and suppose the given length of fencing is L. We then have (8) 2x + y = L. The area inclosed is J. = xy, which from (8) becomes A = x(iL-2x)=Lx-2 x^. Plotting this function, we have the parabola in Fig. 57. We desire to find the value of x corresponding to the vertex V of this parabola, for this gives the greatest value to A. The slope m of the tangent is given by the equation m = L — ix, and this is zero (tangent horizontal at V) when x = \ L. ZZ^ For this value oi x, y = ^ L. The maximum area is therefore obtained when the width is one half of the length. The maximum area F.G. 57 is i L^ square units. 116 MATHEMATICAL ANALYSIS [IV, § 80 Example 2. Three streets intersect so as to inclose a triangular lot ABC. The frontage of the lot on BC is 180 ft. and the point A is 90 ft. back of BC. A rectangular building is to be constructed on this lot so as to face BC. What are the dimensions of the ground plan which will give the maximum floor area? In Fig. 58 we have drawn the lot ABC and have indi- cated by dotted lines two extreme plans. The ground plan sought must be somewhere between these two extremes. To deter- mine its dimensions we proceed as follows : Let X and y be the length of the sides of the ground plan. The floor area (neglecting the thickness of the walls) is (9) A = xy. In order to express A, for which we seek a maximum, in terms of x alone, we now proceed to express y in terms of x. The triangles ABC and ^ilifiV" are similar. (Why?) Hence we have This gives whence (10) From (9) and (10) we obtain MN^^LA BC DA x 90- (Why ?) 180 90 ' y = - i a; -H 90. A = 90 X - I x^. This expresses the floor .area as a function of the side x. The slope of the tangent to the graph is given by mz=90 - X and this slope is zero when x = 90, which in turn gives (by (10)) y = 45, and therefore A = 4050. The maximum area is then 4050 sq. ft. and this is obtained by making the building 90 ft. long and 45 ft. deep. Draw the graph of the function A = 90x-^ ^ x'^. IV, §80] QUADRATIC FUNCTIONS 117 We may note that in both of these examples, the function of which the maximum was sought was obtained as a function of two variables. The conditions of the problem, however, made it possible to express one of these variables in terms of the other and thus to obtain the desired function as a quadratic function of one variable, whereupon the solution was readily effected. The difficulty in this type of problem is usually in connection with the elimination of all but one of the variables. To solve such a problem it is necessary to keep in mind the following steps. (1) Decide, and express in words, of what function a maxi- mum or a minimum value is to be found. (2) Express this function algebraically. (3) If this expression contains more than one variable, use the conditions of the problem to find a relation or relations connecting these variables. (4) By means of the relation or relations found, eliminate all but one of the variables from the function of which a maxi- mum or minimum value is sought. (5) Proceed with the algebraic computation. EXERCISES 1. The number 100 is separated into two parts such that the product of the parts is a maximum. Find the parts and the corresponding product. -^ns. 50, 50, 2500. Is it possible to separate 100 into two parts such that the product of the corresponding parts is a minimum ? Explain. 2. Prove that the rectangle of given perimeter which has the maxi- mum area is a square. 3. Find the greatest rectangular area that can be inclosed by 100 yd. of fence. 4. Separate 20 into two parts such that the sum of their squares will be a minimum. 118 MATHEMATICAL ANALYSIS [IV, § 80 5. A man desires to build a shed against the back of his house, the ground plan to be a rec- tangle. The roof is to be 1 ft. higher in the back than in the front (see the adjoined figure). He has on hand enough siding to cover 253 sq. ft. Allowing 18 sq. ft. for a door and assuming that the height from the ground to the lowest part of the roof is 8 ft., what should be the dimen- sions of the ground plan in order to get the greatest floor area ? 6. An underground conduit is to be built, the cross section of which is to have the shape of a rec- tangle surmounted by a semicircle. If the cost of the masonry is proportional to the perimeter, and if the perimeter is 30 ft., what should be the dimensions of the cross section in order that the conduit will have a maximum capacity ? 7. The same problem as in Ex. 6 with the perimeter of the cross section given as a ft. 8. Determine the greatest rectangle that can be inscribed in a given acute angled triangle whose base is 2 5 and whose altitude is 2 a. *9. In the corner of a field bounded by two perpendicular roads a spring is situated 8 chains from one road and 6 chains from the other. How should a straight path be run by this spring and across the corner so as to cut off as little of the field as possible ? Ans. 12 and 16 chains from the corner. 81. Table of Squares. We have stated that the more important functions have been tabulated (§ 28). The function x^ is one of these. Tables of squares are very helpful in shortening computation. A comparatively rapid method of constructing such a table is given in Ex. 2 below. Here we may make use of our knowledge of the function x^ to see that for a sufhciently small interval in such a table, we are justified in using linear interpolation (§ 56). Indeed we have seen that * The function whose minimum is sought is not in this case quadratic An approximate solution may be obtained graphically. The solution may be computed by finding the slope of the graph from the definition of slope. IV, §81] QUADRATIC FUNCTIONS 119 in a sufficiently small neighborhood of any point on the graph of y = x^, the graph differs as little as we please from a straight 2 line. (See Fig. 45.) For example, if in the second table on p. 100 we confine ourselves to only three-place .94 .884 accuracy, we find that the successive differences in .95 .903 , p . , . -, , g22 the function are almost proportional to the corre- .97 .941 sponding differences in the variable. We give in .960 the adjoined table an extract from the table men- tioned. From this table we may conclude that (.953)2 =.909. This result is accurate only to the third decimal place. EXERCISES 1. Find by interpolation from the above table the following : (.954)2; (.981)2; (9.66)2; (9.89)2. 2. Compute by actual multiplication the squares of all the integers from 31 to 40. This method of computing a table of squares becomes very laborious. Write the results obtained from your computation in a column, and write op- posite each pair of successive squares their difference as shown in the adjoined beginning of such a table. These differences are called the flrst differences of the table. Do you ob- serve any regularity in the formation of these differences? Prove in general the law here suggested. [Hint. Consider the difference between A;2 and (k + 1)^.] Use this law to construct a table of squares from 41 to 100. 3. If the successive differences of the Jirst differences are formed, we obtain the so-called second differences. Prove that in a table of squares of successive integers the successive second differences are all equal to 2. The first differences, therefore, have the character of a linear function. Hence show how to compute the exact value of (.S2.6)2 from the value of (32)2 and (33)2^ This process is known as quadratic interpolation. X a:2 Difference 31 32 33 961 1024 1089 63 65 34 35 36 120 MATHEMATICAL ANALYSIS [IV, § 82 III. QUADRATIC EQUATIONS 82. Definitions. An equation of the form ax^ -\- bx + c = 0^ where a, b, and c are constants and a =^ 0, is called a quadratic equation. A value of x which when substituted in the equation ax"^ -\- bx -{- c = makes both members identical is called a root Example 1. ^s 3 a root of the equation 2x^— 5x-\-6 = 0? Substituting 3 for x, we find 2.3- - 5.3 +Q = 9 and not 0. Therefore 3 is not a root. Example 2. Determine k so that one root of 2 kx- — 3 ar + 5 = shall be 1. Since 1 is to be a root, we have 2 A: — 3 -|- 6 = 0, or k =— 1. The e( [nation then becomes — 2 ic- — 3 a; + 5 = 0. 83. The Roots of ax^-^bx-\-c =0. It follows from § 79 that the equation ax^ -f- te + c = may be written in the form 5 \2 52 _ 4 ofc V 2 ay 4a provided a ^ 0. Dividing by a and solving for (x + 6/(2 a)), we have ^ ^^_^ 1^ + 2^>=^U? ac or 6 , V62-4ac 2a 2a hence (11) b ± V62-4ac 2a We have shown up to this point that if ax^ -{-bx-\- c has the value 0, then x must have one of the values given in equa- tion (11). We need still to prove the converse : If — bH-Vb2 — 4flc — 6 — V62 — 4ac X = X_r or x = , 2a ' 2a IV, § 82] QUADRATIC FUNCTIONS 121 then ax^ -\-hx-\- c will have the value 0. This can be done by- substituting the values of x in turn in the expression ax^-^-hx-^-c and simplifying the resulting expressions.''^ The last part of this proof is essential. We know that the converse of a true theorem may be false. t The first part of our discussion proved that no other values of x than those given by (11) will satisfy the equation aoi^ -\-bx-\- c = 0, but it did not prove that either of these values does satisfy the given equation. Equations (11) maybe used as a formula for solving a quadratic equation. Thus, solving 2 aj2 _ 5 aj _ 13 = where a = 2, 6 = — 5, c = — 13, we have or 5±V25-4(2)(-13) x_ 4 5 ± Vl29 Solution by Factoring. If the factors of a quadratic equation may be found readily, one may proceed as in the following example. Example. Solve x"^ - Sx -\- 2 =0. This equation may be written in the form (a;-2)(x-l) = 0. Therefore, x-2 = or X- -1 = 0, i.e. x = 2 or x=l. Why? See § 48. * The converse can be proved at present only if &2 _ 4 ac is not negative. Why ? t Thus the converse of the true statement, " A horse is an animal," would be the false statement, " An animal is a horse." 122 MATHEMATICAL ANALYSIS [IV, § 83 EXERCISES Determine whether the roots of the following equations are as stated. 1. x2_ 5a;4-6 =0 ;2, 3. 4. 2^2 _ 5^ + 3 = O; 1, - 2. 2. a;2 + 5 x - 6 = ; 1, 2. 5. a;2 - 7 = ; V7, - V7. 3. x2 - 12 a: + 30 = ; 5, 6. 6. 7 a:2 _ 2 x + 51 := ; 0, 1. In the following equations determine k so that the number beside the equation is a root. Find the other root. 7. a;2 + 2 Ax - 5 = ; 1. 8. A:x2 - 6 x + A;2 - 1 = ; 0. Ans. A; = 2 ; other root = — 5. 9. kx'^ — 6 A:x + 11 = fc ; 2. Solve the following equations by means of the formula and also by completing the square : 10. (ax + &)2 = 6 X. 15. sx2 + to - p = 0. 11. (x-5)(7x-3) = 12. jg x2 (3 x + 2)2 ^.^ 12 y + 5 y^ - 5 ^ g . ■ 4 1 7 3 * 17. 3(5x2 -10)+ 2x- 5=0. 13. x2 -f A-x - c?x2 + /i = 0. 18. x2 + (p- g)x-i)g = 0. 14. wP':r^ -\- m{n—p)x — mp = 0. Solve the following equations by factoring : 19. x2 _ 8 x + 15 = 0. 22. 3 x2 - 17 X + 10 = 0. 20. x2 - 14 X + 48 = 0. 23. • 5 X + 14 = x2. 21. 12 - X - x2 = 0. 24. a&x2 + a2x + h'^x + a6 = 0. 25. A cross-country squad ran 6 miles at a certain constant rate and then returned at a rate 5 miles less per hour. They were 50 minutes longer in returning than in going. At what rate did they run ? Ans. 9 miles per hour. 26. When a single row of rivets is used to join together two boiler plates, the distance p between the centers of the rivets is given by the formula ^ = 0.56^+(?, where t is the thickness of the plate and d is the diameter of a rivet hole in inches. In a certain make of boiler the rivets are 1 inch apart and the plate is \ inch thick. Find the diameter of the rivet holes. 27. How high is a box that is 6 ft. long, 2 ft. wide, and has a diagonal 8 ft. in length ? IV, § 84] QUADRATIC FUNCTIONS 123 28. The effective area JS" of a chimney is given by the formula E = A — O.GVA^ Avhere A is the measured area. Find the measured area when the effective area is 25 square feet. 29. Two men can row 12 miles downstream and back again in 5 hours. If the current is flowing at the rate of 1 mile per hour, how fast can the men row in still water ? 30. Find the outer radius of a hollow spherical shell an inch thick whose volume is 76 tt/S cubic inches. [Hint. The volume of a sphere is 47rrY3.] 84. Graphic Solution. Example. Solve x^ — 4.x + 3=0 graphically. Let us plot the graphs oi y == x^, y = 4:X— 3 with reference to the same set of axes (Fig. 59). We see that the two graphs intersect in two points, the coordinates of which satisfy both equations. Therefore the abscissas of these points are values of x which make the right-hand members equal, i.e., for which 0^2 = 4a; -3 or a;2_ 4a; 4.3^:0. The roots are seen to be 1 and 3. If the line and the parabola were tangent, what would you say concerning the roots ? If the line and parabola do not meet, what would you say concerning the roots ? This problem may be solved graphically in an entirely different way. We will plot the curve y =zx^ — 4:X-{-3 (Fig. 60). The abscissas of the points where this graph meets the a?-axis are roots of the original equation. Why ? Describe the roots if the parabola touches the ic-axis. What would you say concerning the roots if the parabola did not meet the ic-axis ? Fig. 59 Fig. 60 124 MATHEMATICAL ANALYSIS [IV, § 85 85. General Theorems. 1. If r is a root of the equation ax"^ 4- 6a; + c = 0, then x — r is a factor of ax"^ -{-hx -\- c. Dividing ax"^ + hx -[- chy x — r, we obtain : X — r\ ax^ -\-hx + c \ ax -\-{}j -\- ar) ax"^ — arx {b -h ar) X -\- c (b 4- ar) x — {b + ar) r c -{- br -\- ar"^' Therefore ax"^ -\-bx -\- c= [cix + {b -\- ar)']\x — r] + c + 5r + ar^. But, by hypothesis, r is a root ; therefore, ar"^ + br + c = 0\ hence ax^ -\-bx -{- G = \_ax -\-{b -\- ar)'][_x — ?-]. 2. Prove that ifx — r is a factor of ax"^ -\-bx -\- c, then x = r is a root of ax^ -\- bx + c = 0. 3. Prove that if the expression ax^ -\- bx -\- c is divided by X — r, the remainder is ar^ -\- br -\- c. The Discriminant of the Quadratic. In § 83 we saw that the roots of the equation ax^ -{- bx -\- c = are -6+V62-4ac ^^^ -b -V62_4(^c The expression under the radical, namely, &2 _ 4 qc^ is called the discriminant of the equation, because it enables us to dis- criminate as to the nature of the roots. From geometric con- siderations we know that a quadratic equation with real coefficients a, b, c may have either two real distinct roots, two real equal roots, or no real roots at all. The above formula enables us to see the same truth algebraically. If b^ — 4:ac = 0, we say that there are two real and equal roots, each being — 6/2 a. If 52 _ 4 fl(< -> 0, there are two real and unequal roots. IV, § 85] QUADRATIC FUNCTIONS 125 If 62 _ 4 ac < 0, there are no real roots. The roots of such an equation are called imaginary or complex. The properties of such numbers will be discussed fully in Chap. XVIII. If the discriminant b^ — A ac is a perfect square and the coefficients a, b, c are rational numbers, then the roots are rational. By finding the value of the discriminant we may determine the nature of the roots of the quadratic without solving the equation. Thus, in the equation Sx'^-i-ix — 3 = 0, the dis- criminant is 52 and we conclude that the roots are real, un- equal and irrational. Eelation of Roots to Coefficients. Let the roots of the equation ax- -f- 6a; -f c = be denoted by /•, and ra . That is, let — 64-V62 — 4ac „ 1 —b — Vb"^ — 4 ac and ?'2 = 2a 2a By addition we have -5.|-V 52_4qe-5-V6'^-4ac _ 2b_ b '*^ + '*^- "" 2a ~ 2a~ a By multiplication we have _ r(- b)-^¥ - 4 ac\\{- b) + ^¥ - 4 ac'] '"''''- 4a^ _ 6^^ — 6^ 4- 4 ac _ c ~ 4a2 a Therefore, if we write the quadratic equation in the form x^-\-^x + -=0, a a the above results may be expressed as follows : In a quadratic equation in ivhicJi the coefficient of the x^ term is unity, (/) the sum of the roots is equal to the coefficient of x with the sign changed; (ii) the product of the roots is equal to the constant term. 126 MATHEMATICAL ANALYSIS [IV, § 85 EXERCISES Solve graphically (two ways) each of the following equations : 1. 2x2 -|-5x- 3:^0. 3. 12-a; = a;2. 5. 4-x2 = 0. 2. a;2-8x + 15=0. 4. 2x2-3x-5=0. 6. 4 + x2 = 0. Form the equations with the following roots : 7. 4,-5. Ans. x2+x-20:=0. 9. 2+V5, 2-V5. 8. V7, - V7. 10. c + 3 &, c - 3 b. 11. What is the remainder when 3x2 — 2x + 5=0 jg divided by x-3? byx + 2? byx-1? by-x+1? [Hint : Use 3, § 85.] Determine, the nature of the roots of the following equations : 12. 7x2_5x = 6. 14. 2 1/2 + 3 2/ + 24 = 0. 13. 2x = 7-3x2. 15. 9x2 = 4x-5. Determine k so that the following equations shall have equal roots. [Hint : Place b'^ — iac equal to zero.] 16. kx^ - 6 x + 3 = 0. Ans. k=S. 18. x^ + 2 {1 -^ k)x -{- k^ = 0. 17. 3iK2-4^•x+ 2 = 0. 19. 2Arx2+(5A: + 2)x + 4^ + l=0. 20. Determine the limits on k so that equations 16-19 shall have their roots real and unequal ; imaginary and unequal. 21. If X is real, show that ~ must lie between and 1. a;2-5x + 9 11 22. A party of students hired a coach for ^ 12, but three of the students failed to contribute towards the expense, whereupon each of the others had to pay 20 cents more. How many students were in the party ? 23. Cox's formula for the flow of water in a long horizontal pipe con- nected with the bottom of a reservoir is Hd^ iv^ + ^^v-2 ' L 1200 where H is the depth of the water in the reservoir in feet, d the diameter of the pipe in inches, L the length of the pipe in feet, and v the velocity of the water in feet per second. If a reservoir contains 49 ft. of water, find the velocity of the water in a 5-inch pipe that is 1000 ft. long. 24. It takes two pipes 24 minutes to fill a certain reservoir. The larger pipe can fill it in 20 minutes less time than the smaller. How long does it take each pipe to fill the reservoir ? Ans. 60 min. ; 40 rain. 25. Prove algebraically and geometrically that if h^ — ^ac<:iQ^ the value of the function ax^ + &x -f c is positive for all (real) values of x, if a > ; and negative for all (real) values of x, if a < 0. IV, § 86] QUADRATIC FUNCTIONS 127 86. Equations involving Radicals. The method of solving problems involving radicals will be illustrated by some examples. Example 1. Solve Va; + ^ - 2 (a; — 1)= 0. Transposing the second term to the right-hand member gives V»T2 = 2(x- 1). Squaring, x-\-2=^4.x'^-Sx-\-4:, or 4a.'2 -9 a; -f- 2 = 0. Whence , a; = 2, or i. Do both these values satisfy the equation? We have shown that, if VicH-2 — 2 (ic — 1)=0, then x = 2 or J. But we cannot conclude conversely, that if a; = 2 or \, then VxT2 - 2 (it- - 1) = 0. In fact, if we substitute the values of x found in the original equation, we find that a; = 2 is a root ; but a; = ^ is not. Example 2. Solve the equation Va; + 8 + Vx + 3 = 5 Vx. Squaring both sides, we find a; -f 8 + 2Va;2-f llaj-f 24 + x + 3 = 25a;, or 2 Va;2+ 11 a; + 24 = 23 x - 11 ; whence squaring, collecting terms, dividing by 25, we have 21 a;2 - 22 a; + 1 = ; therefore, a; = 1 or ^V- What are the roots ? EXERCISES Solve each of the following equations : 1. Vic - 2 - 3 = 0. 4. - \/4 X - 3 - Vx + 1 = 1. 2. >/x-2-(-3 = 0. 5. v'x+5+ Vx+10=v2x + 15. Ans. No roots. $. Vx4- b + Vx-\- a = V2x+a + b. 3. Vx~+2-VxTT = - 1. 7. V2 X + 6 - Vx + 4 = Vx - 4. Ans. 2. 8. Vx+3 - V4 x + 1 = V2 - 8 x. 128 MATHEMATICAL ANALYSIS [IV, § 86 MISCELLANEOUS EXERCISES Determine the condition existing among a, 6, c so that the equation ax'^ + bx + c = shall have : 1. One root double the other. Ans. 2b^ = 9 ac. 2. The roots reciprocals of each other. Ans. a = c. 3. One root three times the other. 4. One root n times the other. 5. One root zero. Ans. c = 0. 6. One root equal to 1 ; 2 ; 3 ; n. 7. The roots numerically equal but opposite in sign. A7is. 6 = 0. 8. Find the area of the largest rectangle that can be inscribed in a triangle whose base is 20 inches and whose altitude is 15 inches, if one side of the rectangle is along the base of the triangle. 9. Separate twenty into two parts such that the product of half of one part by a quarter of the other shall be a maximum. 10. Solve the equation y^ — Sy"^ -^ \b = 0. [Hint : Let y^ = x.] 11. Solve the equation fa; + - 1 + T^ + ^1 — 12 = 0. 12. Solve the equation x^ + 8 x + 3 V^ + 8x4-2= 8. 13. Solve the equation — ^^^ = — • x + l x-^ 12 14. Find k so that the roots of {k + 2)x^ — 2 A;x + 1 = are equal. 15. Without solving, determine the sum and product of the roots of each of the following equations : (a) 2.r2_7x-3 = 0. , (c) 4ic2 _ 3^: + 1 =0. (6) x2 - 4 X + 2 = 0. (d) 2 a;2 + .3 x + 4 = 0. 16. Determine k so that the sum of the roots of the equation 2x2+(A-- l)x+(3A: -7) = 0is4. Ans. k = -l. 17. Determine k so that the product of the roots of the equation (2 A; - 1) x2 + (A; + 3) » + (A;2 - 2 A; + 1) = is 2. CHAPTER V THE CUBIC FUNCTION. THE FUNCTION a^ 87. The General Cubic Function ax^ -f- bx"^ -\- ex -\- d. Hav- ing discussed in the last chapter the general quadratic function ax^ -{-bx -\- c, we now turn our attention to the general algebraic function of the third degree, i.e. the general cubic function. It is of the form OQ^ -\- bx^ -\- ex + dj a=^ 0. 88. The Function ^. We begin with the consideration of the function y = a^. A brief tabular representation of this function is given below. We note that the values of x^ for negative values of x are the same in absolute value as those for the corresponding positive values, but negative. If the corresponding points are plotted with respect to a pair of rec- tangular axes, we obtain Fig. 61. The change Ay in y due to a change Ax in x is calculated as follows, where x^ and y^ are any pair of corresponding values of x and y : (1) y^ + Ay = x^ + 3 a;i2 . Aa; H- 3 x^^'^ + Aa^. K 129 X a;3 0.00 .5 0.12 1.0 1.00 1.5 3.36 2.0 8.00 2.5 15.62 3.0 27.00 ^j^ x in \ ' ' --3^'Alh --Jr-S^S--^ Fig. 61 130 Since this gives (2) MATHEMATICAL ANALYSIS 2/1 = ^ly IV, § 88 A?/ = (3 x^^ + 3 x^^x + Aa;2) Aa;. We can now conclude that as Aa; approaches zero, Ay also approaches zero ; i.e. the function is continuous for all values of X. From (2) we obtain (3) ^ = 3a;i2-j-3a^iAa; + Aa-2 Ao; (if Ax z^ 0). As Aaj (and, therefore, also A?/) approaches 0, this change ratio approaches 3 x^. The slope m of the graph at the point {x^ , y^ is, therefore. (4) m = ^x^\ This slope is positive for all values of Xy^ except a?! = 0. Why ? The function is therefore an increasing function for all values of x except a? = 0, i.e. at the origin, where the graph of the function is tangent to the X-axis. The graph is exhibited in Fig. 62, where we have drawn at certain points the tangents to the graph by means of (4) in order to insure greater accuracy. - i^ :: J::::: 1 -L- .- ± JL :: -"M : 3^::::: di T ::^ : :t::::: «' , i '-% "r ' jL Jrf T '1- e> 9 7: '. :. ::::?_>:3 ^-4i5- / -^ _. it > "L — ":::: :: :: : ■-!«!: ::::: :: Fig. 62 89. The Functions ax^ and a{x - hf + ft. From the results of the last article and the general principles previously established, we conclude that the graph of the function y = ay? is obtained from that of 2/ = a^ by stretching or contracting all the ordinates in the ratio | a | : 1, according as \a\ is greater than 1 or less than 1, and in case a is negative reversing the V, § 89] CUBIC FUNCTIONS 131 signs of all the ordinates (Fig. 63).* Explain the reason for this result. The function y == a{x ~ Tif -{- k may be written in the form y — 7c = a{x — hy. Its graph is accordingly (§ 78) obtained from that of ?/ = aa^ by sliding the latter graph through a distance and in a direc- y=a(z-h)3 + k Fig. 64 tion represented by the motion from (0, 0) to (7 = {h, 7c). Ex- plain the reason for this (Fig. 64). The slope of y = aa^ at the point (xi, y{) is 3 axi^. The slope of a (a; - hy+ k at the point {x^, y{) is Sa{xi- hy. The proof of these statements is left as an exercise. EXERCISES 1. From the graph of the function y — cc^, determine the volume of a cube whose edge is 0.5 in. ; 0.5 ft. ; 3 f t. ; 1.5 cm. 2. Find the equation of the tangent and the normal to the curve y =.3^ at the point (2, 8) ; (_ 1, _ 1) ; (0, 0) ; (- 2, - 8). 3. Draw each of the curves y= -x^, y=4i x^, y=S(x-iy, y=2(x-\-iy * For example, if the unit on the y-scale of Fig. 62 be doubled (i.e. made equal to the a;-scale) while the curve is left unaltered, the graph there given will be the graph of y = i cc^. 132 MATHEMATICAL ANALYSIS [V, § 89 4. Show that the slope oiy = — ax^ at the point (xi, 2/1) is — 3 axi^. 6. Discuss the locus of y = — xK 6. Discuss the locus of y =— ax^ if a is positive and greater than 1 ; less than 1. Show that the same curve will serve as the graph for all values of a > if the units on the axes are properly chosen. 90. The Addition of a Term mx. Shearing Motion. If to an expression in x defining a function, a term of the form mx be added, the effect on the graph is readily described in terms of a type of motion that is important in mechanics. For ex- ample, let us take the function a^ and investigate the effect produced upon the graph by adding the term —Sx. The graphs of y = a^ and y=:—Sx are dravm in Fig. 65. The graph of y = a^ — Sx is then obtained by adding ^ /" \-A=r?--5a the corresponding ordinates of the former / ^ graphs. The addition of these two func- tions is obtained graphically by sliding the ordinate of each point on y = a^ vertically up or down until the base of that ordinate meets the graph of y =: — Sx. If we think of the ordinates of 2/ = 25^ as attached to the avaxis and constrained to remain vertical, the graph of y = a^ will become the graph of y = a^ — 3x if the a>axis is rotated about the origin until it coincides with the line y = — Sx. The resulting graph of 2/ = 0? — 3 a; is, of course, to be interpreted as drawn with ref- erence to the original a>-axis. The motion just described, whereby y = 0^ is transformed into y = a^ — Sxj is called a shearing motion or a shear with respect to y = — Sx. In general, if the term mx is added to aa^, the graph of the function cta^ -{-mx is obtained by subjecting the graph of cue* to a shear with respect to the line y = mx. .If a and m have the same signs, the effect is in the direction of straightening V, § 90] CUBIC FUNCTIONS 133 the graph ; if a and m have different signs, the effect is in the direction of emphasizing the curvature. These effects can be produced by drawing the original figure on the edges of a pack of cards, or on the edges of a book, and then shifting the cards (or sheets of paper) as shown in Fig. 66. EXERCISES Draw the graph of the following functions, making use of the shear : 1. y = Sx'^ + x. 5. y=:ic8 + x-l. 2. ?/ = x2 + X. B. y=-x^ + x-^2. 3. y=-x^-x. 7. y = x^-l. 4. y = - 2 x3 + 4 x. 8. y = x2 - 4 X. 9. Show that y = mx is the equation of the tangent to the curve = x^ 4- mx at the origin. 134 MATHEMATICAL ANALYSIS [V, § 91 91. The Functions a(x - hf + m{x - h)-\- k and ax^ + bx^ -{- cx + d. We have seen that the graph of (6) y = ax^ -f mx has one of the following forms (Fig. 67) according to the signs of a and m. If such a graph is subjected to a parallel motion which Fig. 67 carries the origin to the point {h, k), the equation of the graph in its new position is (§ 78) (6) y — k = a{x — hy -\-m{x— h), which when expanded takes the form (7) 2/ = «^' — 3 ahx^ + (3 ah^ -\-m)x— ah^ — mh -f k. This is of the general form (8) y =z ao? -^ hx^ -[- ex + d. V, § 93] CUBIC FUNCTIONS 135 Moreover, it includes for all values of 7i, k, m all the equations of the general form (8). For (7) and (8) will be identical if (9) - 3 aA = 6, 3 a/i2 -(- m = c, - ah^ - mh -{- k = d. The first of these equations determines h(a^O) ; h being known, the second equation determines m; m and h being known, the third equation determines k. ■ We may conclude then that the graph of any function of the form (8) has one of the shapes given in Fig. 67, but with the origin moved to a point {h, k) given by the equations (9). In order to draw the graph of a function of the form (8) we could first transform (8) into the form (6) and then pro- ceed as in § 90. It is more expeditious, however, to proceed more directly by making use of the slope of the function (8) and our knowledge of what shapes may be expected. 92. The Slope of y = ax^ + bx^ -\-cx + d. The change Ay in y due to a change Ax in the function y = aa^ -\- bx^ -{-ex -j- d, when a; = aji , is Ay = (3 axi^ -{- 2 bx^ -\- c -\- 3 ax^Ax +- bAx -f aAx'^) Ax. This equation shows that the graph is continuous. Why? When Ax approaches 0, the change ratio Ay /Ax approaches the slope m, by definition. This gives, 93. To draw the Graph of y = aj^ -h bx^ -\- ex -{- d. We shall illustrate by means of two examples the method of draw- ing the graph of a cubic function. 136 MATHEMATICAL ANALYSIS [V, § 93 Example 1. Draw the graph oi y = a^ + x'^ — x + 2. The slope m at the point (a;i, yi) is (§ 92) m = S Xi^ -\- 2 Xi — 1. We seek first the points (if such exist) at which the tangent is horizontal, i.e. where m = 0. The roots of the equation m = 0, viz. 3a;i2 + 2a;] -1 = are Xi = — 1 and Xi = -|. The slope is therefore at the points (-1,3) and (^, If). We now compute a table of corresponding values of x, y, m for values of x on both sides of and between x — 1 and x = ^. Such a table and the corresponding figure are given below. X y m -3 -2 - 13 20 7 -1 3 2 -1 1 If 3 4 2 12 15 Fia. 68 Example 2.* Draw the graph of y = — a^ — Sx^ — Sx-^-l. The slope at the point where x = Xi is (3a;i2 + 6aJi + 5). m Since the roots of the equation 3 iCi^ -f 6 cci + 5 = are im- aginary, the graph has no horizontal tangents and the slope m is negative at every point. We accordingly make a table of values and construct the graph (Fig. 69). V, § 94] CUBIC FUNCTIONS 137 X y m -8 16 -15 -2 7 - 5 -1 4 - 2 1 - 5 1 -8 -14 Fia. 69 94. Maxima and Minima. We extend our definition of maximum and minimum given in § 75 as follows : A value of x for which a function stops increasing and begins to decrease is said to correspond to a maximum of the function ; a value of x for which the function stops decreasing and begins to increase is said to correspond to a minimum of the function. Thus in Ex. 1, § 93 the value x = —1 corre- sponds to the maximum 3 of the function ; the value a; = -J corresponds to the minimum ^ of the function.* EXERCISES Draw the following curves and locate in each case the maximum and minimum points if there are any : 6. y = oc^ ■\- X + 1. 7. y = oc^_. 8. y = x^ — X. 9. y =z x^ + 2 x"^ -h X. 1. y = x^ + x^. 2. v=f-5|!-a.., 3. y^X^- — -2x-\-\' 4. y = x^-x'^-5x+2. 6. y = 2a;8 4-^-4x + l. 10. y cfi -x^ -\-x—l. * Note that a maximum of a function does not mean the greatest value a function can assume. In Ex. 1, § 93, the value of the function is greater when x = 2 than when x = —l. It does mean a value of the function which is greater than the values in the immediate neighborhood. 138 MATHEMATICAL ANALYSIS [V, § 9; 95. Geometric Problems in Maxima and Minima. The theory just explained has an important application in solving problems in maxima and minima, i.e. the determination of the largest or the smallest value a magnitude may have which satisfies certain given conditions. As we saw in § 80, the first step is to express the magnitude in question algebraically. If the resulting expression contains more than one variable, other conditions always will be given which will be sufficient to express all of the variables in terms of one of them. When the magnitude in question is expressed in terms of one variable, we can proceed as in § 92 to find any maximum or minimum values which there may be. Example 1. Find the greatest cylinder that can be cut from a given right circular cone, whose height is equal to the diameter of its base. Fig. 70 Let li be the given height of the cone and x and y the un- known dimensions of the cylinder (Fig. 70). Then the volume V of the cylinder is equal to irx^y. But from similar triangles we have V, § 95] CUBIC FUNCTIONS 139 Therefore, whence Now F= TTxXh -2x) = Trhx^ - 2 no^, m = 2 irhx — 6 ttx^. The roots of the equation m = are a; = and x =h/S. It is left as an exercise to draw the graph of the function V = Trhx" - 2 TTCl^ and show that the value x = h/3 corresponds to the maximum of the function, i.e. to y = h/S. Therefore the maximum volume of the cylinder is obtained when the altitude is equal to the radius of the base. The maximum volume is 7r^'/27 or 12/27 of the volume of the cone. EXERCISES 1. A square piece of tin, the length of whose side is a, has a small square cut from each corner and the sides are bent up to form a box. Determine the side of the square cut away so that the box shall have the maximum cubical contents, Ans. a/6. 2. Assuming that the strength of a beam with rectangular cross section varies directly as the breadth and as the square of the depth, what are the dimensions of the strongest beam that can be sawed from a round log whose diameter is d. Ans. Depth = Vf d. 3. Find the right circular cylinder of greatest volume that can be in- scribed in a right circular cone of altitude h and base radius r. Ans. Radius of the base of the cylinder equals f r. 4. Equal squares are cut from each corner of a rectangular piece of tin 30 inches by 14 inches. Find the side of this square so that the re- maining piece of tin will form a box of maximum contents. 5. Show that the maximum and minimum points on the curve y = x^ — ax + 6 (a > 0) are at equal distances from the y-axis. 6. Find the maximum volume of a right cone with a given slant height L. 140 MATHEMATICAL ANALYSIS [V, §96 96. The Power Function. The functions x" and l/a;», where n is any positive integer, are called power functions of x. The curves y = .t'» (Fig. 71) are known as parabolic, while the curves y = l/iC* (Fig. 72) are known as hyperbolic. The curves of the parabolic type possess the property that they all pass through the point (0, 0) and the point (1, 1). r Y T H " \ 4 . ili/L \ *f / 4- ^ \ a >. \ H -g >%- .- '■ Tr ^ ._, . V , V / / \ u > -8 C \/ \- -> V f \ .. / / \ 4) C*J / \ >' / N ill \ w 1^ 4i \ / m s, / / fit [ s / /// 1 ^ »^ A ^ ^ -> ' 'V _![ / ^ =; Z' a / j5 X _j '/ / // ' / f / / J 'V ^14. - PoLJv, J. i n, , rv' JS - - -tsi ^-^ 5 V Iclk. "^ 1 ?/=LrcZl__ y^ -%] [ ^ ti 1 , ^i72r,3;£ )j"l )- - - 5 i or-p Ji ■;3- i ^3 5 Q J- 1 Fig. 71 The larger the value of n, the greater is the slope of the tan- -^ent at the point (1, 1). The curves of the hyperbolic type all pass through the point (1, 1). As X approaches 0, the corresponding value of y becomes infinite. At a; = the value of y is undefined. As x becomes infinite, the corresponding value oiy approaches 0. V, § 96] CUBIC FUNCTIONS 141 -y _1M. 1 Jv J]J 111 ^ •< ^ ~\h ,>^ ^ il. V^l ^ y jLii-L Hr SpL ~^ j ^\ \ 11 T_^ VjLl "^ ::ix l_ / J4l 1 1 1 \\ [7 \\ \| \t \J vl 1 ^ J ^ V, ■^ / V'^ )^'> f r-lO^ \' ■^& =^ li X y-. m^^ ■^~' --= ^~'*p CT Y) -- rrr gmijj,.^] s f 1 1 "" "— "--L ^^\ _i_ -:k s\ T ly-perbo ■ck ^'S 1 \ 1 1 r/ /»# L -x .^i?:._ fo "has 1,2, -tv/c ^,10 \ ]yh ' 1 X \ r. ) Darts \ ^ I \ -A J. Fig. 72 EXERCISES 1. Draw the curves y =x^\ y = x^ ; y = x^; y = x^. 2. Draw the curves y = l/x; ?/ = l/x^ ; y = 1/x^ ; y = 1/ic*. 3. Prove that the slope of the tangent at the point (1, 1) to the curve y = x2, is 2 ; to the carve y—x^ is 3 ; to the curve y=x^ is 4 ; lo the curve y = x^ is 5. 4. Prove that for every even value of n, the parabolic curves ' ?/ = x" pass through the point (- 1, 1); and that for every odd value of n, they pass through the point (— 1, — 1). 6. Prove that the function x^ is an increasing function for all values of x. 6. Find the equation of the tangent and the normal to y = ar^ at the point (2, 32). 7. Prove that the slope of the curve y = 1/x at the point (xi, yi) Is — 1/xi*. [The curve y — l/x is called a hyperbola.'] 142 MATHEMATICAL ANALYSIS [V, § 96 ace* be obtained from the 8. How can the graph of the function y graph of y = x" if a is positive ? negative ? 9. Find the equation of the tangent and the normal to the curve y = l/a:2 at the point (2, \) . 10. Prove that all hyperbolic curves lie within the shaded regions of the adjoining figure, while all parabolic curves lie in the regions left unshaded. CHAPTER VI THE TRIGONOMETRIC FUNCTIONS 97. The functions we have discussed hitherto, namely, the functions of the form mx i-b, ax^ -^bx -\- c, ax^ + bx"^ + ex -\- dj have all been defined by means of explicit algebraic expres- sions. They are all examples of a very large class of functions known as algebraic functions. We now turn our attention to functions defined in an entirely different way. As we shall see, these functions depend on the size of an angle. They enable us to express completely the relations between the sides and the angles of a triangle, and they are of the greatest practical importance in surveying, engineering, and indeed in all branches of pure and applied mathematics. 98. Directed and General Angles. In elenJentary geometry an angle is usually defined as the figure formed by two half- lines issuing from a point. However, it is often more serviceable to think of an angle as being generated by the rotation in a plane of a half-line OP about the point as a pivot, start- y^ \ ing from the initial position OA and / ^^-^"'^ \ ending at the terminal position OB (Fig. 73). We then say that the line OP has generated the angle AOB. Similarly, if OP rotates from the initial position OB to the terminal position OA, then the angle BOA is said to be generated. Considerations similar to those regarding directed line segments (§ 6) lead us to regard one of 143 144 MATHEMATICAL ANALYSIS [VI, § 98 the above directions of rotation as positive and the other as negative. It is of course quite immaterial which one of the two rotations we regard as positive, but we shall assume from now on, that counterclockwise rotation is positive and clockwise rotation is negative. Still another extension of the notion of angle is desirable. In elementary geometry no angle greater than 360° is considered and seldom one greater than 180°. But from the definition of an angle just given, we see that the revolving line OP may make any number of complete revolutions before coming to rest, and thus the angle generated may be of any magnitude. Angles generated in this way abound in practice and are known as angles of rotation.^ When the rotation generating an angle is to be indicated, it is customary to mark the angle by means of an arrow starting at the initial line and ending at the terminal line. Unless some such device is used, confusion is liable to result. In Fig. 75 e^ 0^ Q 30' 390" 750' lilO FiQ. 75 angles of 30°, 390°, 750°, 1110° are drawn. If the angles were not marked one might take them all to be angles of 30°. 99. Measurement of Angles. For the present, angles will be measured as in geometry, the degree (°) being the unit of measure. A complete revolution is 360°. The other units in this system are the minute ('), of which 60 make a degree, and the second ("), of which 60 make a minute. This system of units is of great antiquity, having been * For example, the minute hand of a clock describes an angle of — 180® in 30 minutes, an angle of — 540° in 90 minutes, and an angle of —720° in 120 minutes. VI, § 101] TRIGONOMETRIC FUNCTIONS 145 used by the Babylonians,* The considerations of the previous article then make it clear that any real number, positive or negative, may represent an angle, the absolute value of the number representing the magnitude of the angle, the sign representing the direction of rotation. 100. Angles in the Four Quadrants. Consider the angle XOF = e, whose vertex O coincides with the origin O of a system of rec- tangular coordinates, and whose initial line OX coincides with the positive -p^Y-. -0^^ Fig. 76 half of the x-axis (Fig. 76). The angle d is then said to be in the first, second, third, or fourth quadrant, according as its terminal line OP is in the first, second, third, or fourth quadrant. 101. Addition and Subtraction of Directed Angles. The meaning to be attached to the sum of two directed angles is analogous to that for the sum of two directed line segments. Let a and h be two half-lines issuing from the same point and let (a&) repre- sent an angle obtained by rotat- ing a half-line from the position a to the position 6. Then if we have two angles (a&) and (6c) with the same vertex 0, the sum (a6)-|-(6c) of the angles is the angle represented by the rotation of a half-line from the position a to the position h and then rotating from the position 6 to the position c. But these two rotations are together equivalent to a single rota- tion from a to c, no matter what the relative positions of a, &, c may have * The terms minutes and seconds are derived from their Latin names, which are partes minutss primse and partes minutx secundss. At present there is a slight tendency among some authors to divide the degree decimally instead of into minutes and seconds. Still other authors use the degree and minute and divide the minutes decimally. Exercises involving both these systems will be found in the text. When the metric system was introduced at the end of the eighteenth century it was proposed to divide the right angle into 100 parts, called grades. The grade was divided into 100 minutes and the minute into 100 sec- onds. This system is used in some European countries, but not at all in America. 146 MATHEMATICAL ANALYSIS [VI, § 101 been. Hence, we have for any three half-lines a, 6, c issuing from a point 0, (1) (a&) + (6c) = (ac), (ab) + (&a) = 0, (ab) = (cb)-(ca). The proof of the last relation is left as an exercise. These relations are analogous to those of § 35 ; but an essential difference must be noted. Given two points A and 5 on a line, we may speak of the directed segment AB. The measure of AB is completely determined when A and B and the unit of measure are given. But if the half-lines a and b are given, the angle (ab) may be any angle generated by a rotation from a to 6. Such angles may be positive or negative and may involve, in addition to the minimum rotation from a to &, any number of complete revolutions. It is to be noted, however, that all possible determi- nations of the angle (ab) differ among themselves only by integral multi- ples of 360°. In other words, if 6 represents the smallest positive measure (in degrees) of an angle from a to b, then any determination of (ab) is given by the relation (ab) = 6 ± n- 360° (n an integer). The equality signs in relations (1) are then to be interpreted as meaning eqtial, except for multiples o/360°. If the position of the half-line h is determined by the angle di which it makes with a given horizontal line OX, and the position of another half-line h is deter- mined by the angle 62 which it makes with OX we have angle from Zi to ^2 = 60 — 61 , except for multiples of 360°. Why ? EXERCISES 1. What angle does the minute hand of a clock describe in 2 hours and 30 minutes ? in 4 hours and 20 minutes ? 2. Suppose that the dial of a clock is transparent so that it may be read from both sides. Two persons stationed on opposite sides of the dial ob- serve the motion of the minute hand. In what respect will the angles de- scribed by the minute hand as seen by the two persons differ ? 3. In what quadrants are the following angles : 87° ? 135°? - 326° ? 540°? 1500°? -270°? 4. In what quadrant is 6/2 if ^ is a positive angle less than 360° and in the second quadrant ? third quadrant ? fourth quadrant ? 5. By means of a protractor construct 27°-}- 85° -f (—30°) -f 20°-|- (— 45°). 6. By means of a protractor construct — 130° -\- 56° — 24°. VI, § 102] TRIGONOMETRIC FUNCTIONS 147 102. The Sine, Cosine, and Tangent of an Angle. We may now define three of the functions referred to in § 97. To this end let 9 = XOP (Fig. 80) be any directed angle, and let O X ^ Fig. 80 w T^ US establish a system of rectangular coordinates in the plane of the angle such that the initial side OX of the angle is the positive half of the i»-axis, the vertex being at the origin and the i/-axis being in the usual position with respect to the aj-axis. Let the units on the two axes be equal. Finally, let P be any point other than on the terminal side of the angle 6, and let its coordinates^ be («, y). The directed segment OP = r is called the distance of P and is always chosen posi- tive. The coordinates x and y are positive or negative accord- ing to the conventions previously adopted. We then define ordinate of P y The sine of 9 The cosine of d = The tangent of Q = distance of P abscissa of P distance of P ordinate of P provided x =^ 0,* abscissa of P These functions are usually written in the abbreviated forms sin 0, cos 6, tan 6, respectively ; but they are read as " sine 0/^ " cosine 6/^ " tangent ^." It is very important to notice that the values of these functions are independent of the position of the point P on the terminal line. For let P{x', y') be any other point on this line. Then from the similar right triangles * Prove that x and y cannot be zero simultaneously. 148 MATHEMATICAL ANALYSIS [VI, § 102 xyr* and a;'?/'r' it follows that the ratio of any two sides of the triangle xyr is equal in magnitude and sign to the ratio of the corresponding sides of the triangle xfy'r\ There- fore the values of the functions just defined depend merely on the angle 6. They are one-valued functions of 6 and are called trigonometric functions.^ Since the values of these functions are defined as the ratio of two directed segments, they are abstract numbers. They may be either positive, negative, or zero. Eemembering that r is always positive, we may readily verify that the signs of the three functions are given by the following table. Quadrant Sine Cosine Tangent l' + + + 2 + 3 + 4 103. Values of the Functions for 45°, 135°, 225°, 315°. In each of these cases the triangle xyr is isosceles. Why? Since the trigonometric functions are independent of the position of the point P on the terminal line, we may choose the legs of the right triangle xyr to be of length unity, which ^ j^l P' FiQ. 81 gives the distance OP as V2. Figure 81 shows the four angles * Triangle xyz means the triangle whose sides are x,y,z. t Trigonometric etymologically means relating to the measurement of triangles. The connection of these functions with triangles will appear presently. VI, § 104] TRIGONOMETRIC FUNCTIONS 149 with, all lengths and directions marked. Therefore, sin 45'=-^, V2 sin 135° sin 225^ V2' sin 315° = V2' vr cos 45°= ■" , V2' tan 45° = 1, cos 135°= ^ , V2 tan 135° - - 1, cos 225°= ^ , V2 . tan 225° = 1, cos315°=-i:„ tan 315° = - 1. V2^ 104. Values of the Functions for 30°, 150°, 210°, 330°. From geometry we know that if one angle of a right triangle con- tains 30°, then the hypotenuse is double the shorter leg, which is opposite the 30° angle. Hence if we choose the shorter leg (ordinate) as 1, the hypotenuse (distance) is 2, Jft' £n<^:^ v^ •VJ .^^' 330 Fig. 82 and the other leg (abscissa) is V3. Figure 82 shows angles of 30°, 150°, 210°, 330° with all lengths and directions marked. Hence we have sin 30° =i cos 30° = Y, tan 30°=-^, V3 sin 150° =i cos 150° = -^, tan 150° = -—, V3 sin 210° = - 1 2' cos 210°=-^, tan210°=-J:^, V3 sin 330°=- 1 2' cos 330° = ^^, ton 330° = --^. V3 150 MATHEMATICAL ANALYSIS [VI, § 105 105. Values of the Functions for 60°, 120°, 240°, 300°. It is left as an exercise to construct these angles and to prove that sin 60° = sin 120° = sin 240° = - sin 300° = - 106. Applications. The angle which a line from the eye to an object makes with a horizontal line in the same vertical plane is called an angle of elevation or an angle of depression. V3 2 ' cos 60° = L 2 tan 60°= a/3, V3 > 2 cos 120° = -1, tan 120° -^ - V3, V3 2 ' COS 240°=- ^, tan 240° = v'3. V3 2 ' cos 300° =|, 2 tan300° = -V3. C^ Horizontal ^.^^^"'""" Fia. 83 Horizontal ^^^ according as the object is above or below the eye of the observer (Fig. 83). Such angles occur in many examples. Example 1. A man wishing to know the distance between two points A and B on opposite sides of a pond, locates a point C on the land (Fig. 84) such that ^C = 200 rd., angle G — Z^\ and angle B = 90°. Find the distance AB. ^=sin(7. (Why?) AG AB=AC sin a = 200 . sin 30° = 200 . I = 100 rd. Example 2. Two men stationed at points A and C 800 yd. apart and in the same vertical plane with a balloon B, observe simultaneously the angle of elevation of the balloon to be 30° and 45° respectively. Find the height of the balloon. Solution : Fia. 84 VI, § 106] TRIGONOMETRIC FUNCTIONS 151 Solution : Denote the height of the balloon DB by y, and let DC = x then AD = 800 - x. tan 45° = 1, we have 1 = ^ X Since and since tan 30° = 1/ V3, we have V3 800 - X Therefore oc = y and 800 — x = y \' 8. 800 Solving these equations for y, we have V3+ 1 292.8 yd. EXERCISES 1. In what quadrants is the sine positive ? cosine negative ? tangent positive ? cosine positive ? tangent negative ? sine negative ? 2. In what quadrant does an angle lie if (a) its sine is positive and its cosine is negative ? (6) its tangent is negative and its cosine is 'positive ? (c) its sine is negative and its cosine is positive ? (d) its cosine is positive and its tangent is positive ? 3. Which of the following is the greater and why : sin 49° or cos 49° ? sin 35° or cos 35° ? 4. If 6 is situated between 0° and 360°, how many degrees are there in 6 if tan ^ = 1 ? Answer the similar question for sin = ^ ; tan ^ = - 1. • 5. Does sin 60° = 2 • sin 30° ? Does tan 60° = 2 • tan 30° ? AYhat can you say about the truth of the equality sin 2 ^ = 2 sin ^ ? 6. The Washington Monument is 555 ft. high. At a certain place in the plane of its base, the angle of elevation of the top is 60°. How far is that place from the foot and from the top of the tower ? 7. A boy \vhose eyes are 5 ft. from the ground stands 200 ft. from a flagstaff. From his eyes, the angle of elevation of the top is 30°. How high is the flagstaff ? 152 MATHEMATICAL ANALYSIS [VI, § 106 8. A tree 38 ft. high casts a shadow 88 ft. long. What is the angle of elevation of the top of the tree as seen from the end of the shadow ? How far is it from the end of the shadow to the top of the tree ? 9. From the top of a tower 100 ft. high, the angle of depression of two stones, which are in a direction due east and in the plane of the base, are 45° and 30° respectively. How far apart are the stones ? ^ns. 100(V3-l)=73.2ft. 10. Find the area of the isosceles triangle in which the equal sides 10 inches in length Include an angle of 120°. Ans. 25 VS = 43.3 sq. in. 11. Is the formula sin2 ^ = 2 sin ^cos ^ true when = 30°? 60°? 120^^^^ ? 12. From a figure prove that sin 117° = cos 27°. 13. Find the tangent of the angle which the line joining the points (^1) yi)i and (X2, Vi) makes with the aj-axis, assuming the units on the two axes to be equal. Compare your answer with the definition of slope in §§ 50 and 53. 14. Determine whether each of the following formulas is true when e = 30°, 60°, 150°, 210° : 1 + tan2 d 1 + cos'^ d 1 1 tan2 d sin2 0' sin2 + cos2 = 1. 16. Let Pi(a:i, yi) and P2(a;2, 2/2) be any two points the distance be- tween which is r (the units on the axes being equal). If is the angle that the line P1P2 makes with the cc-axis, prove that X2-X1 J y2 - yi ^ 2 r. cos 6 sin 6 107. Computation of the Value of One Trigonometric Function from that of Another. Example 1. Given that sin = f , find the 6y\ hs5 values of the other functions. XB I _J__1^^ Since sin 6 is positive, it follows that d is an angle in the first or in the second quad- rant. Moreover, since the value of the sine is I, then y = Z - k and r = 6 •% where k is any positive constant different from zero. (Why ?) It is, of course, immaterial what positive value we assign to A;, so we shall assign the Fia. 86 VI, § 108] TRIGONOMETRIC FUNCTIONS 153 value 1. We know, however, that the abscissa, ordinate, and distance are connected by the relation x"^ + y^ = r^, and hence it follows that X = ± 4. Fig. 86 is then self-explanatory. Hence we have, for the first quadrant, sin ^=f, cos^=|, and tan^=|; for the second quadrant, sin ^ = |, cos ^ = — I, tan 5 = — |. Example 2. Given that sin d = f^ and that tan d is negative, find the other trigonometric functions of the angle 6. Since sin 6 is positive and tan 6 is negative, 6 must be in the second quadrant. We can, therefore, con- struct the angle (Fig. 87), and we obtain sin ^ = j\, :^1_ Fig. 87 108. Computation for Any Angle. Tables. The values of the trigonometric functions of any angle may be computed by the graphic method. For example, let us find the trigonometric functions of 35°. We first construct on square ruled paper, by means of a protractor, an angle of 35° and choose a point P on the ter- minal line so that OF shall equal 100 units. Then from the figure we find that 0M= 82 units iind MP = 57 units. Fig. 88 Therefore sin 35° = ^ = 0.57, cos 35° = -^\\ = 0.82, tan 35° = || = 0.70. The tangent may be found more readily if we start by tak- ing OA — 100 units and then measure AB. In this case, AB = 70 units and hence tan 35° = t% = ^•'^^• It is at once evident that the graphic method, although :it IIHi :: HI mill :: i; :: lHlllllljlllll iW m Iks: m ;;;::: :: :: :: iiiiiiiisi;:;: ip^ :s!ii iZHSt 't fPffl iii ±i + iSSSSSSSSSSSSSS ::::! ■■■■■ :s: .4+1 'i T1 i; !■■■■■ sSsssssss :::: 1 ;| 11 i. ^ iiillll i±; mt k 1 iii? lliiii ;; + : +£ u:i:i::i:i:K:i :±i: i ilss :: :; :l::Kypi::::: -1 1 g ■4- it i ii 1 1 == II llil ::: ■:i 1 s i| i: Iwif S== --" + + 1 1 illf ±: Sill! i 1 ^ 1 5^; :: :::: iS*M i Piffl 10 no 30 40 80 90 too 154 MATHEMATICAL ANALYSIS [VI, § 108 10 Fig. 20 30 iO SO CO 70 60 90 100 89. — Graphical Table of Trigonometric Functions VI, § 108] TRIGONOMETRIC FUNCTIONS 155 simple, gives only an approximate result. However, the values of these functions have been computed accurately by methods beyond the scope of this book. The results have been put in tabular form and are known as tables of natural trigonometric functions. These tables with an explanation of their use will be found in any good set of mathematical tables.* In order to solve several of the following exercises it is necessary to make use of such tables. Figure 89 makes it possible to read off the sine, cosine, or tangent of any angle between 0° and 90° with a fair degree of accuracy. The figure is self-explanatory. Its use is illustrated in some of the following exercises. EXERCISES Find the other trigonometric functions of the angle 6 when 1. tan^ = -3. 3. cos ^= if. 6. sin^ = f. 2. sin^=-|. 4. tan^=4. 6. cos^=-^ 7. sin ^ = I and cos 6 is negative. ■ _, 8. tan 6 = 2 and sin 6 is negative. 9. sin e = — ;^ and tan 6 is positive. 10. cos ^ = f and tan 6 is negative. 11. Can 0.6 and 0.8 be the sine and cosine, respectively, of one and the same angle ? Can 0.5 and 0.9 ? Ans. Yes ; no. 12. Is there an angle whose sine is 2 ? Explain. 13. Determine graphically the functions of 20°, 38°, 70°, 110°. Check your results by the tables of natural functions. U. From Fig, 89, find values of the following : sin 10°, cos 50°, tan 40°, sin 80°, tan 70°, cos 32°, tan 14°, sin 14°. 15. A tower stands on the shore of a river 200 ft. wide. The angle of elevation of the top of the tower from the point on the other shore exactly opposite to the tower is such that its sine is |. Find the height of the tower. * See, for example, The Macmillan Tables, which will be referred to in connection with this book. 156 MATHEMATICAL ANALYSIS [VI, § 108 16. From a ship's masthead 160 feet above the water the angle of de- pression of a boat is such that the tangent of this angle* is ^j. Find tlie distance from the boat to the ship. Ans. 640 yards. 17. A certain railroad rises 6 inches for every 10 feet of track. What angle does the track make with the horizontal ? 18. On opposite shores of a lake are two flagstaffs A and B. Per- pendicular to the line AB and along one shore, a line BC = 1200 ft. is measured. The angle ACB is observed to be 40° 20'. Find the distance between the two flagstaffs. 19. The angle of ascent of a road is 8°. If a man walks a mile up the road, how many feet has he risen ? 20. How far from the foot of a tower 150 feet high must an observer, 6 ft. high, stand so that the angle of elevation of its top may be 23°. 5 ? 21. From the top of a tower the angle of depression of a stone in the plane of the base is 40° 20'. What is the angle of depression of the stone from a point halfway down the tower ? 22. The altitude of an isosceles triangle is 24 feet and each of the equal angles contain 40° 20'. Find the lengths of the sides and area of the triangle. 23. A flagstaff 21 feet high stands on the top of a cliff. From a point on the level with the base of the cliff, the angles of elevation of the top and bottom of the flagstaff are observed. Denoting these angles by a and /3 respectively, find the height of the cliff in case sin a ■=. ^ and cos/3 = H. Ans. 76 feet. 24. A man wishes to find the height of a tower CB which stands on a horizontal plane. From a point A on this plane he finds the angle of ele- vation of the top to be such that sin CAB = f . From a point A' which is on the line AC and 100 feet nearer the tower, he finds the angle of elevation of the top to be such that tan CA'B — f . Find the height of the . tower. 25. Find the radius of the inscribed and circumscribed circle of a regu- lar pentagon whose side is 14 feet. 26. If a chord of a circle is two thirds of the radius, how large an angle at the center does the chord subtend ? 27. A boy standing a feet behind and opposite the middle of a football goal observes the angle of elevation of the nearer crossbar to be a, and the angle of elevation of the farther crossbar to be p. Prove that the length of the field is a [tan a — tan /3]/tan /3. VI, § 109] TRIGONOMETRIC FUNCTIONS 157 109. The Sine Function. Let us trace in a general way the variation of the function sin ^ as ^ increases from 0° to 360°. For this purpose it will be convenient to think of the distance r as constant, from which it follows that the locus of P is a circle. When = 0°, the point P lies on the x-axis and hence the ordinate is 0, i.e. sin 0° = 0/r = 0. As ^ increases to 90°, the ordinate increases until 90° is reached, when it becomes equal to r. Therefore, sin 90° = r/r = 1. As ^ p^^ ^ increases from 90° to 180°, the ordinate de- creases until 180° is reached, when it becomes 0. Therefore sin 180° = 0/r = 0. . As ^ increases from 180° to 270°, the ordi- nate of P continually decreases algebraically and reaches its smallest algebraic value when = 270°. In this position the ordinate is — r and sin 270° = — r/r = — 1. When enters the fourth quadrant, the ordinate of P increases (algebraically) until the angle reaches 360°, when the ordinate becomes 0. Hence, sin 360° = 0. It then appears that : as 6 increases from 0° to 90°, sin increases from to 1 ; as 6 increases from 90° to 180°, sin decreases from 1 to ; as 6 increases from 180° to 270°, sin 6 decreases from to — 1 ; as 6 increases from 270° to 360°, sin $ increases from — 1 to 0. It is evident that the function sin 6 repeats its values in the same order no matter how many times the point P moves around the circle. We express this fact by saying that the function sin 6 is periodic and has a period of 360°. In symbols this is expressed by the equation sin [6 + n . 360°] = sin 9, where n is any positive or negative integer. The variation of the function sin 6 is well shown by its 158 MATHEMATICAL ANALYSIS [VI, § 109 graph. To construct this graph proceed as follows : Take a system of rectangular axes and construct a circle of unit radius Fig. 91 with its center on the aj-axis (Fig. 91). Let angle XOP = 0. Then the values of sin 6 for certain values of 6 are shown in the unit circle as the ordinates of the end of the radius drawn at an angle 0. d 30° 45° 60° 90° ... sin^ M,Pi M2P2 MsPs M,P, Now let the number of degrees in ^ be represented by dis- tances measured along OX. At a distance that represents 30° erect a perpendicular equal in length to sin 30° ; at a distance that represents 60° erect one equal in length to sin 60°, etc. Through the points 0, Pi, P^,-" draw a smooth curve'; this curve is the graph of the function sin 6. If from any point P on this graph a perpendicular PQ is drawn to the a;-axis, then QP represents the sine of the angle represented by the segment OQ. Since the function is periodic, the complete graph extends indefinitely in both directions from the origin (Fig. 92). Y VI, § 110] TRIGONOMETRIC FUNCTIONS 159 110. The Cosine Function. By arguments similar to those used in the case of the sine function we may show that : as 6 increases from 0°to 90°, the cos decreases from 1 to ; as $ increases from 90° to 180°, the cos 9 decreases from to — 1 ; as increases from 180° to 270°, the cos 6 increases from — 1 to ; as increases from 270° to 360°, the cos increases from to 1. The graph of the" function is readily constructed by a method Fig. 93 similar to that used in case of the sine function. This is illustrated in Fig. 93. The complete graph of the cosine function, like that of the sine function, will extend indefinitely from the origin in both 7 ^^ /^ \ AA Vy '' \ / \-^ -i ^=cosx Fig. 94 directions (Fig. 94). Moreover cos 0, like sin 0, is periodic and has a period of 360°, i.e. cos [6 + n . 360°] = cos e, where n is any positive or negative integer. 160 MATHEMATICAL ANALYSIS [VI, § 111 Fig. 95 111. The Tangent Function. In order to trace the varia- tion of the tangent function, consider a circle of unit radius with its center at the origin of a system of rectangular axes (Fig. 95). Then construct the tangent to this circle at the point M{1, 0) and let P denote any point on this tangent line. If angle MOP = 0, we have tan 6 = MP/OM = MP /I — MP, i.e. the line MP represents tana Now when 6 = 0°, MP is 0, i.e. tan 0° is 0. As the angle 6 increases, tan 6 increases. As 6 ap]3roaches 90° as a limit, MP becomes infinite, i.e. tan 6 becomes larger than any number whatever. At 90° the tangent is undefined. It is sometimes convenient to express this fact by writing tan 90° = 00. However we must remember that this is not a definition for tan 90°, for oo is not a number. This is merely a short way of saying that as $ approaches 90°, tan becomes infinite and that at 90° tan 6 is undefined. See § 36. Thus far we have assumed 6 to be an acute angle approaching 90° as a limit. Now let us start with 6 as an obtuse angle and let it decrease towards 90° as a limit. In Fig. 96 the line MP' (which is here negative in direction) represents tan 9. Arguing precisely as we did before, it is seen that as the angle 6 approaches 90° as a limit, tan 6 again increases in magnitude beyond all bounds, i.e. becomes infinite, remaining, however, always negative. Fig. 96 VI, § 111] TRIGONOMETRIC FUNCTIONS 161 We then have the following results. , (1) When 6 is acute and increases toward' 90° as a limit, tan always remains positive but becomes infinite. At 90° tan is undefined. (2) When 6 is obtuse and decreases towards 90° as a limit, tan 6 always remains negative but becomes infinite. At 90° tan 6 is undefined. It is left as an exercise to finish tracing the variation of the tangent function as 6 varies from 90° to 360°. Note that tan 270°, like tan 90°, is undefined. In fact tan n • 90° is unde- fined, if n is any odd integer. 360° X Fia. 97 To construct the graph of the function tan 9 we proceed along lines similar to those used in constructing the graph of sin e and cos 0. The following table together with Fig. 97 illustrates the method. d 0° 30° 45° 60° 90° 120° 136° 150° 180° 210° tan^ MPi MP2 MPz undefined MPi MP^ MP^ 3/P7=0 MPi 162 MATHEMATICAL ANALYSIS [VI, § 111 It is important to notice that tan 6y like sin 9 and cos 6j is periodic, but its period is 180°. That is tan (6 + 1 • i8o°)= tan 6, where n is any positive or negative integer. EXERCISES 1. What is meant by the period of a trigonometric function ? 2. What is the period of sin 6 ? cos 6 ? tan d ? 3. Is sin d defined for all angles ? cos d ? 4. Explain why tan 6 is undefined for certain angles. Name four angles for which it is undefined. Are there any others ? 5. Is sin (^ + 360^) = sin d ? 6. Is sin {e + 180°) =sin^? 7. Is tan {d + 180°) = tan ^ ? 8. Is tan {6 + 360°) = tan ? Draw the graphs of the following functions and explain how from the graph you can tell the period of the function : 9. sin^. 11. tan^. 13. -^• cos d 10. cosd. 12. — — 14. ^ . sin d tan 6 Verify the following statements : 16. sin 90° + sin 270° = 0. 18. cos 180° + sin 180° = - 1. 16. cos 90° + sin 0° = 0. 19. tan 360° + cos 360° = 1. 17. tan 180° + cos 180°=- 1. 20. cos90°-f tan 180°-sin270° = l. 21. Draw the graphs of the functions sin ^, cos 6, tan ^, making use of a table of natural functions. See p. 638. 22. Draw the curves y = 2 sin ^ ; y = 2 cos ; y = 2 tan 6. 23. Draw the curve j/ = sin + cos 6. 24. From the graphs determine values of d for which sin ^ = ^ ; sin 6 = 1 ; tan <? = 1 ; cos = ^ ; cos ^ = 1. VI, § 112] TRIGONOMETRIC FUNCTIONS 163 Fig. 98 112. Polar Coordinates. It is convenient at this point to introduce a new way of locating the position of a point in a plane, and of representing the graph of a function. To this end (Fig, 98) let OA be a directed line in the plane which we shall call the initial line or the polar axis. This line is usually drawn horizontally and directed to the right. The point is called the pole or the origin. Let P be any point in the plane and draw the line OP. The position of P is then located completely if we know the angle ^OP=^and the dis- tance OP=:p. The two numbers (p, 9), called respectively the radius vector and the vectorial angle, are known as the polar coordinates of the point P. In Fig. 98 we have represented a case in which and f> are both positive. Either ot p or both may be negative under the following conventions. The angle is positive or negative according to the direction of its rotation, as in § 98. The positive direction on OP is the direction from along the terminal side of the angle 0, i.e., it is the direction into which OA is rotated by a rotation through the angle 0. With these conventions a point P whose polar coordinates (/), 0) are given is completely de- termined. Figure 99 shows points whose polar coordinates are (2, 30°), (-2, 30°), (2, -30°), and (-2, — 30°). It will be noted that, if p is positive, P is on the terminal side of $, while if p is negative, P is on the terminal side produced through 0. On the other hand, a given point P has an unlimited number of polar coordinates (p, 6). Even if we confine ourselves to angles i-s,-so') {2,30') (-s,so ) i2y-30°) Fig. 99 164 MATHEMATICAL ANALYSIS [VI, § 112 in absolute value less than 360°, a point lias in general /owr dif- ferent sets of polar coordinates. Fig. 100 shows that the same (e.so'). point P may be designated by any one of the pairs of values (2, 30°), (2, - 330°), (- 2, 210°), and (- 2, - 150°). EXERCISES 1. Locate the points whose polar coordinates have the following values : (4, 30°), (-2, 45<^), (-3, -60°), (2, -160°), (3, -90°), (2, 180°), (-2, 0°), (0, 90°), (-2, 180°), (- 3, 270°). 2. For each of the points in Ex. 1, give all other sets of polar coordi- nates for which 6 is in absolute value less than 360°. 3. What exceptions are there to the statement " 6 being confined to angles in absolute value less than 360°, every point has four and only four distinct sets of polar coordinates " ? 4. Where are all the points for which ^ is a given constant ? 5. Where are all the points for which p is a given constant ? 113. Graphs in Polar Coordinates. Polar coordinates may be used to represent the graph of a given function, in a way quite similar to that in the case of rectangular coordinates. Fig. 101 gives an example in which the idea of polar coor- dinates is used in practice. In this example the ^-scale rep- resents time, the p-scale represents temper ature.*^ Some forms of self-recording hygrometers employ the same idea. * It will be noted that in this example the radius vector is measured along a circular arc instead of along a straight line. This is due to the mechanical (construction of the instrument. Cf. footnote, p. 9. The fundamental idea is, nevertheless, that of polar coordinates. VI, § 1131 TRIGONOMETRIC FUNCTIONS 165 ' In plotting the graph of a function in polar coordinates we proceed as in the case of rectangular coordinates. A table of Fig. 101 corresponding values of the variable 6 and the function p is 166 MATHEMATICAL ANALYSIS [VI, § 113 constructed. Each such pair of values is then plotted as a point, and a curve drawn through these points. Example. Plot in polar coordinates the graph of p = sin 6. We ob- tain the table below. Figure 102 exhibits the corresponding points, with (1.90) e p = sin ^ 0° .00 30^ .50 46° .71 60° .87 90° 1.00 120° .87 135° .71 150° .50 180° .00 210° - .50 225° - .71 240° - .87 270° - 1.00 300° - .87 315° - .71 330° - .50 360° .00 a curve drawn through them. Observe that each point serves to represent two pairs of corresponding values. Thus the pairs (^, 30°) and ( ~ i, 210°) are represented by the same point. This curve suggests a circle, of diame- ter unity, tangent to the polar axis at the origin. 114. The Graph of sin 6 and cos 6 in Polar Coordinates. We may now prove : The graph, in polar coordinates, of the function p — sin $ is a circle of diameter unity, tangent to the polar axis at the origin. Let P (p, 6) be any point on such a circle (Fig. 103). Then, for any value 6 in the first quadrant OA 1 or p = sin ^. VI, § 114] TRIGONOMETRIC FUNCTIONS 167 Conversely, if p = sin 6, the point P is on the circle. Why ? A similar proof, which is left as an exercise, may be given when 6 is in the second, third, or fourth quadrants (Fig, 104). Similarly, we may prove : FiQ. 1(M Hie graph of p = GO8 in polar coordinates is a circle of diameter unity, passing through the pole and having its center on the polar axis. The proof of this statement is left as an exercise. See Figs. 105, 106. On account of their simplicity, the polar graphs of sin $ and cos 6 are very serviceable. It is for this reason that we have FiQ. 105 Fig. 106 introduced them at this point. Polar coordinates will be dis- cussed again, particularly in Chapter XIV, and incidentally in other chapters. 168 MATHEMATICAL ANALYSIS [VI, § 114 EXERCISES 1. From Fig. 101, find the temperature at 9 p.m. on Tuesday ; at 3 p.m. on Monday. When was the temperature a maximum ? a minimum ? 2. Plot in polar coordinates the graph representing the variation in temperature given in Ex. 1, p. 16. 3. Plot the graph in polar coordinates of the function p = tan d. Why is this graph not convenient to represent the function tan 6 ? 4. Prove that the graph, in polar coordinates, of /> = a cos ^ is a circle of diameter a, passing through the origin and w^ith its center on the polar axis. 5. Prove a theorem regarding the graph of p = a sin d, 115. Other Trigonometric Functions. The reciprocals of the sine, the cosine, and the tangent of any angle are called, respectively, the cosecant, the secant, and the cotangent of that angle. Thus, cosecant = distance of P ^ r ( i^^^ ^ o). ordinate of P y ^^ ^ ^ , /J distance of P r , ., , ^^.^ secant 6 = — — : = - (provided x^O), abscissa of P x . , /, abscissa of P a? , • i •, , r^. cotangent 6 = — = - (provided y^O). ordinate oi P y These functions are written esc 6, sec 0, ctn d. From the definitions follow directly the relations csce = ^i-, sece = -, ctn0 = sin 6' COS0' tan 9' or esc ^ • sin ^ = 1, sec ^ • cos ^ = 1, ctn • tan ^ = 1. To the above functions may be added versed sine (written versin), the coversed sine (written coversin), and the external secant (written VI, § 116] TRIGONOMETRIC FUNCTIONS 169 exsec), which are defined by the equations versin 6 = 1 — coa 6, coversin e = 1 — sin 6, and exsec 6 = sec d — 1. It is left as an exercise to trace the variation of esc By sec 6, ctn 0, as varies from 0° to 360°. Be careful to note tliat ctn 0°, ctn 180°, esc 0°, esc 180°, sec 90°, sec 270° are undefined. Why? 116. The Representation of the Functions by Lines. We have seen in §§ 109-111, that if we take a unit circle we may represent sin 9, cos 0, and tan by means of lines. We will now extend this representation to include esc 6, sec 6, ctn 6. Fig. 107 Figure 107 shows the functions in a unit circle for an angle 6 in the first quadrant. We have MF = sin e OM=Gos6 AT BS tan^ ctn^ Or=sec^ OS = CSC $ Draw similar figures for angles in each of the other quad- rants. The points may be so labeled that the results given for the first quadrant hold in any quadrant. 170 MATHEMATICAL ANALYSIS [VI, § 117 117. Relations among the Trigonometric Functions. As one might imagine, the six trigonometric functions sine, cosine, tangent, cosecant, secant, cotangent are connected by certain relations. We shall now find some of these relations. From Fig. 80 (§ 102) it is seen that for all cases we have (1) • 0:2 4- 2/2 = ,.2_ If we divide both sides of (1) by r^, we have ^ -f- ^ = 1 (by hypothesis r^0)\ or sin2 + cos'e = l. Dividing both sides by a;^, we have 1 + ^ = ^ iiix^O), x^ x^ Therefore 1 + tan^ 6 = sec2 6. Similarly dividing both sides of (1) by if gives ^ + 1 = S (if.'/^O); or ctn2 e + 1 = csc2 e. Moreover, we have X X COS r and, similarly, ctne=5?i|. sm6 VI, § 118] TRIGONOMETRIC FUNCTIONS 171 118. Identities. By means of the relations just proved any expression containing trigonometric functions may be put into a number of different forms. It is often of the greatest importance to notice that two expressions, although of a different form, are nevertheless identical in value. (See § 47 for the definition of an identity.) The truth of an identity is usually established by reducing both sides, either to the same expression,- or to two expres- sions which we know to be identical. The following examples will illustrate the methods used. Example 1. Prove the relation sec^ d + csc^ d = sec^ d csc^ d. We may write the given equation in the form sec2 d csc2 5, or which reduces to cos'^ d sin^ ?Hl!i±^2sif = sec2ecsc2^, cos2 dsin^d sec2 d csc2 d, cos2 e sin2 d sec2 d csc2 d = sec2 d csc^ 6. Since this is an identity, it follows, by retracing the steps, that the given equality is identically true. Both members of the given equality are undefined for the angles 0°, 90°, 180°, 270°, 360° or any multiples of these angles. cos*^ 6 Example 2. Prove the identity 1 + sin ^ = ■, — -• 1 — sm ^ Since cos^ ^ = 1 — sin2 ^, we may write the given equation in the form 1 + sin = ^ ~ ^^"^ ^ or 1 + sin d = 1 + sin 6. 1 - sin ^ As in Example 1, this shows that the given equality is identically true. The right-hand member has no meaning when sin = 1, while the left- hand member is defined for all angles. We have, therefore, proved that the two members are equal except for the angle 90° or (4 ji -|- 1) 90°, where n is any integer. 172 MATHEMATICAL ANALYSIS [VI, § 118 The formulas of § 117 may be used to solve examples of the type given in § 107. Example 3. Given that sin d = /^ and that tan 6 is negative, find the values of the other trigonometric functions. Since sin^ d + cos^ ^ = 1, it follows that cos ^ = ± ^|, but since tan 6 is negative, 6 lies in the second quadrant and cos 6 must be — ||. More- over, tlie relation tan d = sin 0/ cos 6 gives tan 6 =— j%. The reciprocals of these functions give sec ^ = — |f, esc 6 = ^^-^ ctn ^ = — -^. EXERCISES 1. Define secant of an angle ; cosecant; cotangent. 2. Are there any angles for which the secant is undefined ? If so, what are the angles ? Answer the same questions for cosecant and co- tangent. 3. Define versed sine ; coversed sine. 4. Complete the following formulas : sin20 + cos2 0=? l + tan2^ = ? l+ctn2^=? tan5=? Do these formulas hold for all angles ? 5. In what quadrants is the secant positive ? negative ? the cosecant positive ? negative ? cotangent positive ? negative ? 6. Is there an angle whose tangent is positive and whose cotangent is negative ? 7. In what quadrant is an angle situated if we know that (a) its sine is positive and its cotangent is negative ? (6) its tangent is negative and its secant is positive ? (c) its cotangent is positive and its cosecant is negative ? 8. Express sin^ ^ + cos ^ so that it shall contain no trigonometric function except cos 6. 9. Transform (1 + ctn^ 6) esc so that it shall contain only sin 6. 10. Which of the trigonometric functions are never less than one in absolute value ? 11. For what angles is the following equation true : tan = ctn ? 12. How many degrees are there in when ctn ^ = 1 ? ctn^ = — 1 ? sec ^ = \/2 ? CSC = \/2 ? 13. Determine from a figure the values of the secant, cosecants **»^ cotangent of 30°, 160^, 2W\ 330°. VI, § 118] TRIGONOMETRIC FUNCTIONS 173 14. Determine from a figure the values of the secant, cosecant, and cotangent of 45% 135°, 226% 316°. 16. Determine from a figure the values of the sine, cosine, tangent, secant, cosecant, and cotangent of 60^, 120°, 240°, 300°. 16. Show that the graphs of the function sec e, esc ^, ctn d have the forms indicated in the adjacent figures. Prove the following identities and state for each the exceptional values of the variables, if any, for which one or both members are undefined : 17. cos d tan d = sin d. 18. sin 6 ctn Q = cos 6. 1 + sin g _ cos5 cos 6 19 1 — sin g 20. sin2 6 — cos2 6 = 2 sin2 5-1. 21. ( 1 - sin2 5) csc2 6 = ctn2 ^. 22. tan 6 + ctn 5 = sec 5 esc 0. 23. [x sine-t y cos ey -h [xcosd-y sin 6^ = x^ + ^. CSC 6 24. = cos 0. tan + ctn 26. 1 - ctn* = 2 csc2 - esc* 0. 26. tan2 5-sin2 5 = tan2 5sin2g. 27. 2(1 + sin 0) ( 1 + cos 0) = (1 + sin + cos ey. 28. sin» + cos« 5 = 1—3 sin2 cos2 6. CSC 5 29. _^!5^ + 30. CSC 5 — 1 CSC 5 + 1 1 — tan 9 _ Ctn g - 1 1 + tan 5 ctD 5 -f 1 = 2 sec2 e. 174 MATHEMATICAL ANALYSIS [VI, § 118 31. [1 + tan e + sec e][l + ctn e — esc ^] = 2. 32. (taiid + sec^)2 = LiLSEi. ^ ^ 1 - sin ^ 33. CSC* ^ (1 - cos* d)-2 ctn2 ^ = 1. 34. (tan d — ctn ^)sin ^ cos = 1 - 2 cos^ e. 36. sec g - tan j ^ i _ 2 sec g tan ^ + 2 tan^ d. sec + tan 6 36. ^-^?i«±J^^5j = tanatan/3. ctn a + ctn /3 37. sin (sec d + esc ^) — cos (sec ^ — esc 6) = sec ^ esc d. Find algebraically the other trigonometric functions of the angle when 38. ctn = 4: and sin is negative. 39. sin = I and sec is positive. 40. sec ^ = 2 and tan is negative. 41. CSC ^ = — 5 and ctn is positive. 119. Trigonometric Equations. An identity, as we have seen (§ 47), is an equality between tvro expressions which is satisfied for all values of the variables for which both expres- sions are defined. If the equality is not satisfied for all values of the variables for which each side is defined, it is called a conditional equality, or simply an equation. Thus 1 — cos ^ = is true only if ^ = w • 360°, where n is an integer. To solve a trigonometric equation, i.e. to find the values of for which the equality is true, we usually proceed as follows. 1. Express all the trigonometric functions involved in terms of one trigonometric function of the same angle. 2. Find the value (or values) of this function by ordinary algebraic methods. 3. Find the angles between 0° and 360° which correspond to the values found. These angles are called particular solutions. 4. Give the general solution by adding n • 360°, where n is any integer, to the particular solutions. VI, § 119] TRIGONOMETRIC FUNCTIONS 175 Example 1. Find d when sin 5 = ^. The particular solutions are 30° and 150°. The general solutions are 30° + w . 360°, 150° + n • 360°. Example 2. Solve the equation tan ^ sin ^ — sin ^ = 0. Factoring the expression, we have sin 6 (tan^— 1) = 0. Hence we have sin ^ = 0, or tan ^—1=0. Why ? The particular solutions are therefore 0°, 180°, 45°, 225°. The general solutions are n • 360°, 180° + w • 360°, 45° + n . 360°, 225° + n . 360°. Example 3. Find d when tan 6 + ctn 6 = 2. The given equation may be written tan (? + —^ = 2, tan d or tan2 ^-2 tan ^+1 = 0; therefore (tan ^ - 1)2 = 0, or tan tf = 1. It follows that 6 = 46° or 225° ; or, in general, ^ = 46° + n . 360° or 226° + n • 360°. EXERCISES Give the particular and the general solutions of the following equations : 1. sin d = ^. 9. tan ^ = - 1. 2. sin 5 = - ^. 10. ctn ^ = - 1. 3. cos e= ^. 11. tan 6=1. 4. cos 5 = — ^. 12. ctn ^ = 1. 6. sec 6 = 2. 13. tan2 6 = 3. 6. sec ^ = — 2. 14. sin ^ = 0. 7. CSC 6 = 2. 15. cos ^ = 0. 8. esc ^ = - 2. 16. tan 6 = 0. Solve the following equations giving the particular and the general solutions in each case : 17. sin d = cos 6. Ans. 45°, 226° ; 45° + n • 360°, 226° + n • 360°. 18. tan^^H- 2sec2^ = 6. 19. 6 sin ^ + 2 cos2 6 = 5. Ans. 90° ; 90° + n • 360°. 20. cos2 ^ + 5 sin = 7. 176 MATHEMATICAL ANALYSIS [VI, § 119 21. 4 sin 5 — 3 esc ^ = 0. 22. 2 sin 6 cos^ 5 = sin 5. 23. cos 5 + sec 5 = f . 24. 2 sin = tan d. Ans. Particular solutions : 0'', 180°, 60°, 300°. 25. 3 sin ^ + 2 cos ^ = 2. 26. 2cos2 ^-1 = 1- sin2 e . 120. The Trigonometric Functions of — 9. Draw the angles and — 9, where OP is the terminal line of and OP' is the terminal line of — 0. Figure 108 shows an angle in each of P P X V p7> ^ ^ Fig. 108 X V X the four quadrants. We shall choose OP — OP' and {x, y) as the coordinates of P and (x\ y') as the coordinates of P'. In all four figures Hence x' = ic, y' — — y, r' — T. sin(-^)=^ = :=^ = -sin(9, r r cos (-(9)=^ = -= cos d, r r tan(-^) = ^ = ^:^ = -tand! Also, csc(— ^)= — csc^; sec (— ^)= seed; ctn (-- d)=— ctnd. VI, § 121] TRIGONOMETRIC FUNCTIONS 177 121. The Trigonometric Functions of 90° — 6. Figure 109 represents angles 9 and 90° — 0, when ^ is in each of the f oui quadrants. Let OP be the terminal line of 6 and OP', the Y ^ »' 90'-9 VF X' J ^ X Fio. 109 terminal line of 90° - 6. Take OP' = OP and let (x, y) be the coordinates of P and (a;', y') the coordinate of P. Then in all four j&gures we have ^ = y> y'^^y r' = r. Hence sin(9O°-0) = 4=- = cos^, r r tan (90°-^) — ^ — IL — r r = ^ = ?=ctnd. cos(90°-^) = - = ^=sind, r r y Also, CSC (90° -^)=sec^, sec (90° -(9)=csc^; ctn(90° -^)=tand. Definition. The sine and cosine, the tangent and cotan- gent, the secant and cosecant, are called co-functions of each other. The above results may be stated as follows : Any function of an angle is equal to the corresponding co-function of the com- plementary angle.* *Two angles are said to be complementary if their sum is 90°, regardless of the size of the angles. 178 MATHEMATICAL ANALYSIS [VI, § 122 122. The Trigonometric Functions of 180° — 0. By draw- ing figures as in §§ 120, 121, the following relations may be proved : sin (180° - 0) = sin 0, esc (180° -6)= esc 0, cos (180° -0) = — cos e, sec (180° -6) = - sec 6, tan (180° - (9) = - tan 0, ctn (180° -&)=- ctn 6. The proof is left as an exercise. 123. The Trigonometric Functions of 180° + 6. Similarly, the following relations hold : sin (180° + ^) = - sin 0, esc (180° + 0) = - esc 6, cos (180° 4- ^) = - cos 0, sec (180° + ^) = - sec 6, tan (180° + (9) = tan B, ctn (180° + ^) = ctn d. The proof is left as an exercise. 124. Summary. An inspection of the results of §§ 120-123 shows : 1. Each function of — B or 180° ± B is equal in absolute value (but not always in sign) to the same function of 0, 2. Each function of 90° — B is equal in magnitude and in sign to the corresponding co-function of 6. These principles enable us to find the value of any function of any angle in terms of a function of a positive acute angle (not greater than 45° if desired) as the following examples show. Example 1. Reduce cos 200° to a function of an angle less than 45°. Since 200° is in the second quadrant, cos 200° is negative. Hence cos 200°= - cos 20°. Why ? Example 2. Reduce tan 260° to a function of an angle less than 45°. Since 260° is in the third quadrant, tan 260° is positive. Hence tan 260° = tan 80° = ctn 10° (§ 121). VI, § 124] TRIGONOMETRIC FUNCTIONS 179 EXERCISES Reduce to a function of an angle not greater than 45° : 1. sin 163°. 6. esc 900°. 2. cos(- 110°). 6. ctn (- 1215°). Ans. — cos 70° or — sin 20°. 7, tan 840°. 3. sec (-265°). 8. sin 510°. 4. tan 428°. Find without the use of tables the values of the following functions : 9. cos 570°. 13. cos 150°. 10. sin 330°. 11. tan 390°. W- tan 300°. 12. sin 420°. Reduce the following to functions of positive acute angles : 16. sin 250°. 18. sec (-245°). Ans. — sin 70° or — cos 20°. 19, esc (— 321°). 16. cos 158°. 20. sin 269°. 17. tan (-389°). 21. Prove the following relations from a figure : (a) sin (90° + d) = cos d. (c) sin (180° + 6)= - sin $. cos (90° + e) = - sin d. cos (180° + ^) = - cos 0. tan (90° + d) = - ctn 0. tan (180° + 0) = tan 0. esc (90° + 0) = sec 0. CSC (180° + ^) = - esc ^. sec (90° + ^) = - CSC 0. sec (180° + 0) = -8ec0. Ctn (90° + 0)=- tan 0. ctn (180° -\-0) = ctn 0. (6) sin (180° -0)= sin 0. (d) sin (270° - ^) = - cos 0. cos (180° -0) = - cos 0. cos (270° -0) = - sin 0. tan ( 1 80° - ^) = - tan 0. tan (270° - ^) = ctn 0. CSC (180° - ^) = CSC 0. CSC (270° - ^) = - sec 0, sec (180° -0) = - sec 0. sec (270° -0) = - esc 0. ctn (180° - ^) = - ctn ^. ctn (270° - ^) = tan 0. (e) sin (270° + ^) = - cos ^. cos (270° + ^)= sin ^. tan (270° + 0) = — ctn 0. CSC (270° + ^) = - sec ^. sec (270° + 0)= CSC ^. ctn(270° + ^) = ~ tan^. 180 MATHEMATICAL ANALYSIS [VI, § 125 125. Law of Sines. Consider any triangle ABC with the altitude CD drawn from the vertex C (Fig. 110). I) B D A Fia. 110 In all cases we have sin A= -, sin B =- b a Therefore, dividing, we obtain sin^ a or sin^ 6' a b (2) sin A sin B If the perpendicular were dropped from B, the same argu- ment would give -^ = -^. (3) sin A sin C Combining results (2) and (3) we have a _ b _ c sin A sin 5 sin C* This law is known as the law of sines and may be stated as follows : Amj tivo sides of a triangle are proportional to the sines of the angles opposite these side'i. 126. Law of Cosines. Consider any triangle ABC with the altitude CD drawn from the vertex C (Fig. 111). In Fig. Ill a AD = 6 cos ^ ; CD = 6 sin ^ ; DB = c — beosA. In Fig. Ill b AD = — 6 cos -4 ; CD = 6 sin ^ ; DB = c — b cos A VI, § 127] TRIGONOMETRIC FUNCTIONS In both figures Therefore a2 = c2 — 2 6c cos A-{-¥ cos^ A-\-ll^ sin^ A = c^ — 2 &c cos ^ 4- (cos2 A + sin* A) b% 181 b D B D whence FlQ. Ill a2 = &2 + c2 — 2 6c cos A, Similarly it may be shown that 52 = c2 -f- a2 - 2 ca • cos B, c^ z= 0} ■\- h"^ — 2 ah • cos C. Any one of these similar results is called the law of cosines It may be stated as follows : Tlie square of any side of a triangle is equal to the sum of the squares of the other two sides diminished by tivice the product of these two sides times the cosine of their included angle.* 127. Solution of Triangles. To solve a triangle is to find the parts not given, when certain parts are given. From geometry we know that a triangle is in general determined when three parts of the triangle, one of which is a side, * Of what three theorems in elementary geometry is this the equivalent ? 182 MATHEMATICAL ANALYSIS [VI, § 127 are given.* Eight triangles have already been solved (§ 106 £f.), and we shall now make use of the laws of sines and cosines to solve oblique triangles. TJie methods employed will be illustrated by some examples. It will be found advantageous to construct the triangle to scale, for by so doing one can often detect errors which may have been made. 128. Illustrative Examples. Example 1. Solve the triangle ABC, given ^^^^^ A = 30^^ 20', B = 60° 45', a = 276. ^A Solution : C = 180° - (^ + B)=1S0° - 91° 5' = 88° 55' ; a sin B 276 sin 60= sin J. sin 30= 20 45^ ^ (276) (0.8725) ^^ygQ. I' 0.5050 ' ' also a sin G ^ 276 sin 38° 55' ^ (276) (0.i)998) ^ ^^g ^ Check : It is left as an exercise to show that for these values we have c2 = a2 + &2 _ 2 a6 cos C. Example 2. Solve the triangle ABC, given A = 30°, 6 = 10, a = 6. ^Q^ Constructing the triangle ABC, we see that .^x^so" two triangles AB\ C and AB2C answer the descrip- -^ tion since b>a> altitude CD. Solution : Now Fig. 113 whence But -B2 ?H^ = ^orsin5i=^^Hl^ sin ^ a a Bi = 56.5°. 180° - Bi = 180° - 56.5° = 123.6°, = 0.833, and Ci = 180° - (^ + Bi) = 180° - 86.5° = 93.6°, Ci = 180° -(A + Bi) = 180° - 153.5° = 26.6°. * When two sides and an angle opposite one of them are given, the triangle is not always determined. Why ? VI, § 128] TRIGONOMETRIC FUNCTIONS 183 Now Also cg^ sin (72 ^ oj. c^^ a sin C^ ^ (6) (0.446) ^ g g^^ a sin ^ ' sin ^ 0.600 Ci^ sinCi Qj, asm(7i^ (6)(0.998) ^^^Q3 a sin^' sin J. 0.600 Check : Cy^ = a^ -\- b^ — 2 ah cos (7i. 143.5 = 86+ 100 +(2)(6)(10)(0.061) = 143.3. C22 = a2 4- 62 _ 2 a6 cos O2. 28.62 = 36 + 100-(2)(6)(10)(0;896) - 28.60. Example 3. Solve the triangle ABC, given a= 10, 6=6, 0=40°. Solution : c^ = a^ + 6^ _ 2 a6 cos C j^ = 100 + 36 - ( 120) (0. 766) = 44.08. Therefore c = 6.64. Now \fy^ \V sin .4 = «^^5iZ = ^lOHO-643) ^ ^ ggg c 6.64 ' i.e. -4 = 104.6°. Likewise sin5 = ^«l5_^=(^KM43)^0.581, c 6.64 Check : A + B^- G= 180.0°. Example 4. Solve the triangle ABC when a = 7, 6 = 3, c = 6. From the law of coshies, = -!=- 0.500, 2 6c 2 ^ + c2-62 ^13^0Q^8, 2ac 14 ^2 + 62 - C2 11 n 7QA COS C = = — = 0. 786. 2 a6 14 Therefore ^ ^ 120°, 5 = 21.8°, C = 38.2°. Check -. A^-B+ C= 160.0°. EXERCISES Solve the triangle ABC, given (a) ^ = 30°, 5 = 70°, a = 100; (6)^ = 40°, 5 = 70°, c = 110; (c) A = 45.5°, = 68.6°, 6 = 40 ; (d)J5 = 60.5°, C = 44°20', c = 20 ; 184 MATHEMATICAL ANALYSIS [VI. § 128 (e) a = 30, 6 = 64, 0=50°; (g)a=10, 6 = 12, c = U; (/) 6 = 8, a = 10, O = 60° ; {h) a = 21, 6 = 24, c = 28. 2. Determine the number of solutions of the triangle ABO when (a) A = 30°, 6 = 100, a = 70 (e) A = 30°, 6 = 100, a = 120 (/) A = 106°, 6 = 120, a = 16 ig) A = 90°, 6 = 15, a = 14. (6) A = 30°, 6 = 100, a = 100 (c) ^ = 30°, 6 = 100, a = 50 (d) A = 30°, 6 = 100, a = 40 3. Solve the triangle ABC when (a) ^ = 37° 20', a = 20, 6 = 26 ; (c) ^ = 30°, a = 22, 6 = 34. (6) ^ = 37° 20', a = 40, 6 = 26; 4. In order to find the distance from a point A to z. point JB, a line AC and the angles CAB and ACB were measured and found to be 300 yd., 60° 30', 56° 10' respectively. Find the distance AB. 5. In a parallelogram one side is 40 and one diagonal 90. The angle between the diagonals (opposite the side 40) is 25°. Find the length of the other diagonal and the other side. How many solutions ? 6. Two observers 4 miles apart, facing each other, find that the angles of elevation of a balloon in the same vertical plane with themselves are 60° and 40° respectively. Find the distance from the balloon to each- observer and the height of the balloon. 7. Two stakes A and B are on opposite sides of a stream ; a third stake C is set 100 feet from A, and the angles ACB and CAB are observed to be 40° and 110°, respectively. How far is it from Ato B? 8. The angle between the directions of two forces is 60°. One force is 10 pounds and the resultant of the two forces is 15 pounds. Find the other force.* 9. Resolve a force of 90 pounds into two equal components whose directions make an angle of 60° with each other. 10. An object B is wholly inaccessible and invisible from a certain point A. However, two points C and D on a line with A may be found such that from these points B is visible. If it is found that CD = 800 feet, CA = 120 feet, angle DCB = 70°, angle CDB = 50°, find the length AB. * It is shown in physics that If the line segments AB 3 ~ ^ ^ ^^^ -^^ represent in magnitude and direction two forces r ^^^/ acting at a point A, then the diagonal AJ) of the parallelo- fy^^ / gram ABCD represents both in magnitude and direction A B the resultant of the two given forces. VI, § 128] TRIGONOMETRIC FUNCTIONS 185 11. Given a, 6, A, in the triangle ABC. Show that the number of possible solutions are as follows : [ a < 6 sin ^ no solution, I bsmA<ia<.b two solutions, \a>b I a = 6 sin J one solution. ( a <b no solution, I o > 6 one solution. 12. The diagonals of a parallelogram are 14 and 16 and form an angle of 50°. Find the length of the sides. 13. Resolve a force of magnitude 150 into two components of 100 and 80 and find the angle between these components. 14. It is sometimes desirable in surveying to extend a line such as AB in the adjoining figure. Show that this can be done by means of the broken line ABODE. What measurements are necessary ? 15. Three circles of radii 2, 6, 5 are mutually tangent. Find the angles between their lines of centers. 16. In order to find the distance between two objects A and B on op- posite sides of a house, a station C was chosen, and the distances CA = 500 ft., CB = 200 ft., together with the angle ACB - 65° SO' were measured. Find the distance from A to B. 17. The sides of a field are 10, 8, and 12 rods respectively. Find the angle opposite the longer side. 18. From a tower 80 feet high, two objects, A and B, in the plane of the base are found to have angles of depression of 13° and 10° respec- tively ; the horiz(mtal angle subtended by A and B at the foot C of the tower is 44°. Find the distance from A to B. 186 MATHEMATICAL ANALYSIS [VI, § 129 129. Areas of Oblique Triangles. 1. When two sides and the included angle are given. Denoting the area by S^ we have from geometry G S = ^ch, but ^ = & sin ^ ; therefore (4) S^^cb sin A. Likewise, Fig. 116 S — ^ab sin O and S = ^acsinB. 2. When a side and two adjacent angles are given. Suppose the side a and the adjacent angles B and C to be given. We have just seen that S = ^ac sin B. But from the law of sines we have a sin C Therefore S = SIR A a^ • sin jB • sin C 2 sin A But sin A = sin [180° - (5 + (7)] = sin (5 + C). Therefore ^ _ a'^ sin B sin O ~ 2 sin (5+0)' 3. When the three sides are given. We have seen that S = ^bc sin A. Squaring both sides of this formula and transforming, we have ™ 7)2,.2 7)2^2 S^ = ^ sin2^ = ^(l-cos2^) whence, = |(l4-cos^).|(l-cos^); ^^A 1 b^-\-c^-a' S' = ^ 1-h 2 V 2 6c ;N bcf. b^ + c^-a^ \ J 2[ 2 be J VI, § 130] TRIGONOMETRIC FUNCTIONS 187 4 ' 4 _M-_c4;_a b -\-c—a a~b-\-c g-f 6~c 2*2*2*2' which may be written in the form S^ = s(s-a){s-b)(s-c), where 2s = a + &4-c. Therefore, (5) S = Vs(s-a)(s-6)(s-c). 130. The Radius of the Inscribed Circle. If r is the radius of the inscribed circle, we have from elementary geometry, since s is half the perimeter of the triangle, S = rs ; equating this value of S to that found in equation (5) of the last article and then solving for r, we get, ^J (s-a)(s-b)(s-c) _ ^ 5 EXERCISES Find the area of the triangle ABC, given 1. a = 26, b = 31.4, C = 80° 25'. 4. a = 10, 6 = 7, = 60°. 2. 6 = 24, c = 34 3, ^ = 60° 25'. 5. a = 10, 6 = 12, C=60°. 3. a = 37, 6 = 13, C = 40°. 6. a = 10, 6 = 12, C = 8°. 7. Find the area of a parallelogram in terms of two adjacent sides and the included angle. 8. The base of an isosceles triangle is 20 ft. and the area is 100/ \/3 sq. ft. Find the angles of the triangle. Ans. 30°, 30°, 120°. 9. Find the radius of the inscribed circle of the triangle whose sides are 12, 10, 8. 10. How many acres are there in a triangular field having one of its sides 60 rods in length and the two adjacent angles, respectively, 70° and 60° ? CHAPTER VII TRIGONOMETRIC RELATIONS 131. Radian Measure. In certain kinds of work it is more convenient in measuring angles to use, instead of the degree, a unit called the radian. A radian is defined as the angle at the center of a circle whose subtended arc is equal in length to the radius of the circle (Fig. 117). Therefore, if an angle 6 at the center of a circle of radius r units subtends an arc of s units, the measure of in radians is (1) 0=i. Since the length of the whole circle is 2 irr, it follows that ^^ = 2 TT radians = 360°, r (2) IT radians = 180°. Fig. 117 Therefore, -IQAO 1 radian = ±^ = 57° 17' 45" (approximately). TT It is important to note that the radian * as defined is a con- stant angle, i.e., it is the same for all circles, and can therefore be used as a unit of measure. * The symbol »" is often used to denote radians. Thus 2" stands for 2 radians, tt*- for tt radians, etc. When the angle is expressed in terms of -t (the radian being the unit) , it is customary to omit »•. Thus, when we refer to an angle tt, we mean an angle of tt radians. When the word radian is omitted, it should be mentally supplied in order to avoid the error of supposing ir means 180. Here, as in geometry, IT = ;i. 14159. ... 188 VII, § 132] TRIGONOMETRIC RELATIONS 189 Erom relation (2) it follows that to convert radians into degrees it is only necessary to multiply the number of radians by ISO/V, while to convert degrees into radians we multiply the number of degrees by 7r/180. Thus 45° is 7r/4 radians ; 7r/2 radians is 90°. 132. The Length of Arc of a Circle. From relation (1), § 131, it follows that s=re 5 = re. That is (Fig. 118), if a central angle is measured in radians, and if its intercepted arc and the radius of the circle are measured in terms of the same unit, then ^°' length of arc = radius x central angle in radians. EXERCISES 1. Express the following angles in radians : 25°, 145°, 225°, 300°, 270°, 450°, 1150°. 2. Express in degrees the following angles : IT 7 IT Sir n 57r 4' "T' T'^'^'T* t. A circle has a radius of 20 inches. How many radians are there in an angle at the center subtended by an arc of 25 inches ? How many degrees are there in this same angle ? Ans. f ; 71° 37' approx. 4. Find the radius of a circle in which an arc 12 inches long subtends an angle of 35°. 6. The minute hand of a clock is 4 feet long. How far does its ex- tremity move in 22 minutes ? - 6. In how many hours is a point on the equator carried by the rotation of the earth on its axis through a distance equal to the diameter of the earth ? 7. A train is traveling at the rate of 10 miles per hour .on a curve of half a mile radius. Through what angle has it turned in one minute ? 8. A wheel 10 inches in diameter is belted to a wheel 3 inches in diameter. If the first wheel rotates at the rate of 5 revolutions per minute, at what rate is the second rotating ? How fast must the former rotate in order to produce 6000 revolutions per minute in the latter ? 190 MATHEMATICAL ANALYSIS [VII, § 133 133. Angular Measurement in Artillery Service. The divided circles by means of which the guns of the United States Field Artillery are aimed are graduated neither in degrees nor in radians, but in units called mils. The mil is defined as an angle subtended by an arc of ■g:^^ of the circumference, and is therefore equal to 2_ir_ ^ SAAW ^ 0.00098175 = (0.001 - 0.00001825) radian. 6400 3200 ^ The mil is therefore approximately one thousandth of a radian. (Hence its name.)* Since (§132) length of arc = radius x central angle in radians, it follows that we have approximately T5) li ill A length of arc = x central angle in mils ; 1000 ^ ' i.e. length of arc in yards = (radius in thousands of yards) • (angle in mils). The error here is about 2 %. Example 1. A battery occupies a front of 60 yd. If it is at 5500 yd. range, what angle does it subtend (Fig. 119)? We have, evidently, angle = — = 11 mils. 5.5 Example 2. Indirect Fire, t A battery posted with its right gun at G is to open fire on a battery at a point T, distant 2000 yd. and invisible * To give an idea of the value in mils of certain angles the following has been taken from the Drill Regulations for Field Artillery (1911), p. 164: " Hold the hand vertically, palm outward, arm fully extended to the front. Then the angle subtended by the width of thumb is 40 mils width of first finger at second joint is 40 mils width of second finger at second joint is .... 40 mils width of third finger at second joint is 35 mils width of little finger at second joint is 30 mils width of first, second, and third fingers at second joint is . 115 mils These are average values." t The limits of this text preclude giving more than a single illustration of the problems arising in artillery practice. For other problems the student is referred to the Drill Regulations for Field Artillery (1911), pp. 57, 61, 150-164; and to Andrews, Fundamentals of Military Service, pp. 153-159, from which latter text the above example is taken. VII, § 133] TRIGONOMETRIC RELATIONS 191 from Cr (Fig. 120) . The officer directing the fire takes post at a point B from which both the target T ami a church spire P, distant 3000 yd. from O are visible. B is 100 yd. at the right of the line &T and 120 yd. at the right of the line GP and the officer finds by measurement that the angle P-BT contains 3145 mils. In order to train the gun on the target the gunner must set off the angle PGT on jf rpr the sight of the piece and then move the gun until the spire P is visible through the sight. When this is effected, the gun is aimed at T. Let F and E be the feet of the perpen- diculars from B to G^Tand G^P respectively, and let BV and BP' be the parallels to OT and OP that pass through B. Then, evidently, if the officer at B measures the angle PPT, which would be used instead of angle POT were the gun at B instead of at G^ and determines the angles TBT' = FTB and PBP' = EPB, he can find the angle PGT from the relation PGT=P'BT'= PBT Fig. 120 TBT'- PBP'. Now tan FTB FB TF tan EPB = — . PF Furthermore if FTB and EPB are small angles, i.e., if FB and EB are small compared with OT Sind OP respectively, the radian measure of the angle is approximately equal to the tangent of the angle. Why ? Hence we have FB GT EB OP 100 Therefore FTB - tan FTB = EPB = tan EPB approximately. TBT' = FTB = -^^^ radians 2000 50 mils. PBP' EPB = -^ radians 3000 40 mils. Hence POT = PBT - TBT - PBP* = 3145 - 60 - 40 — 3056 mils, which is the angle to be set off on the sight of the gun. 1^2 MATHEMATICAL ANALYSIS [VII, § 133 Hence for the situation indicated in Fig. 120 we have the following rule : * (]) Measure in mils the angle PBT from the aiming point P to the target T as seen at B. ■ : (2) Measure or estimate the offsets FB and EB in yards, the range G^Tand the distance GB of the aiming point P in thousands of yards. (3) Compute in mils the offset angles by means of the relations TBT = FTB, FBP' = EPB, TBr=—, GT' CrF (4) Then the angle of deflection FGT is equal to the angle FBT diminished by the sum of the offset angles. EXERCISES 1. A battery occupies a front of 80 yd. It is at 5000 yd. range. What angle does it subtend ? 2. In Fig. 120 suppose FBT= 3000 mils, FB = 200 yd., GT = 3000 yd., EB = 150 yd., GP = 4000 yd. Find the number of mils in FGT. 3. A battery at a point G is ordered to take a masked position and be ready to fire on an indicated hostile battery at a point T whose range is known to be 2100 yd. The battery commander finds an observing station B, 200 yd. at the right and on the prolongation of the battery front, and 175 yd. at the right of PGr An aiming point P, 5900 yd. in the rear, is found, and PBT is found to be 2600 mils. Find FGT. 134. Inverse Trigonometric Functions. The equation X = sin y (1) may be read : y is an angle whose sine is equal to a;, a statement which is usually written in the contracted, form y — arc sinflj.f (2) * There are three cases with corresponding rules, depending on whether P is in front of, rear of, or on the flank of G. t Sometimes written y = sin-i.c. Hefe — 1 is not an algebraic exponent, but merely a part of a functional symbol. When we wish to raise sin x to the power — 1, we write (sin a;)-i. VII, § 134] TRIGONOMETRIC RELATIONS 193 For example, x = sin 30° means that x = ^, while y = arc sin^ means that y = 30°, 150°, or in general (n being an integer), 30° + n . 360° ; 150° + n • 360°. Since the sine is never greater than 1 and never less than — 1, it follows that — 1 ^ a; ^ 1. It is evident that there is an unlimited number of values of y = arc sin x for a given value of X in this interval. We shall now define the principal value Arc sin x* of arc sin x^ distinguished from arc sin x by the use of the capital A, to be the numerically smallest angle whose sine is equal to x. This function like arc sin x is defined only for those values of X for whichsk - 1< a; < 1. The difference between arc sin x and Arc sin x is well illus- trated by means of their graph. It is evident that the graph oi y = arc sin x, i.e. X = sin y is simply the sine curve with the role of the x and y axes inter- changed. (See Fig. 121.) Then for every admissible value of x, there is an un- limited number of values of y ; namely, the ordinates of all the points Pi, P2, — , in which a line at a distance x and parallel to the 2/-axis intersects the curve. The single-valued function Arc sin x is repre- sented by the part of the graph between Jlf and iV. Similarly arc cos x, defined as " an angle whose cosine is a;," * Sometimes written Sin-icc, distinguished from sin-^a; by the ase of the capital S. T -. 2t ^ Ps ( sir ir \ n IT i > y N- -i ^ 1 X M G y= arc sin x y= 'Ar (;sinx Fig. 121 194 MATHEMATICAL ANALYSIS [VII, § 134 has an ■anlimited. number of values for every admissible value of a; (— 1^ a; ^ 1). We shall define the principal value Arc cos X as the smallest positive angle whose cosine is x. That is, < Arc cos a? ^ TT. Figure 122 represents the graph of y = arc cos x and the portion of this graph between M and N represents Arc cos x. Similarly we write x = tan y as y,= arc tan X, and in the same way we define the symbols arc ctn x ; arc sec x ; arc esc x. The principal values of all the inverse trigonometric functions are given in the following table. 7 2ir ■) 3jr Ps N ^■J IT ^ \ M -1 IT < 1 X y= arc cos x y=Arc cosz Fig. 122 y = Arc sin x Arc cos X Arc tan x Range of x Range of y X positive X negative -^to^ 2 2 1st Quad. 4th Quad. -l^x^l to T 1st Quad. 2d Quad. all real values 1st Quad. 4th Quad. Arc ctn x Arc sec x Arc CSC X Range of x Range of y X positive X negative all values tOTT 1st Quad. 2d Quad. a; > 1 or a; < - 1 tOTT 1st Quad. 2d Quad. X>l0TX<-l -^ to"^ 2 2 1st Quad. 4th Quad. In so far as is possible we select the principal value of each inverse function, and its range, so that the function is single- valued, continuous, and takes on all possible values. This ob- viously cannot be done for the Arc sec x and for Arc esc y. VII, § 134] TRIGONOMETRIC RELATIONS 195 EXERCISES 1. Explain the difference between arc sin x and Arc sin x. 2. Find the values of the following expressions : (a) Arc sin \. (&) arc sin \. (c) arc tan 1. (d) Arc tan - 1. (e) arc cos I^. (H Arc cos lA. 2 ' 2 3. What is meant by the angle tt ? 7r/4? 4. Through how many radians does the minute hand of a watch turn in 30 minutes ? in one hour ? in one and one half hours ? 5. For what values of x are the following functions defined : (a) arc sin x ? (6) arc cos cc? (c) arc tan x ? id) arc ctn X ? (e) arc sec a; ? (/) arc esc cc ? 6. What is the range of values of the functions : (a) Arc sin x ? (&) Arc cos x ? (c) Arc tan x ? id) Arc ctn x ? (e) Arc sec x ? (/) Arc esc x ? 7. Draw the graph of the functions : (a) arc sin x. (&) arc cos x. (c) arc tan x. (<?) arc ctn X. (e) arc sec x. (/) arc esc x. 8. Find the value of cos (Arc tan |). Hint. Let Arc tan \ = 6. Then tan ^ = f and we wish to find the value of cos e. 9. Find the values of cos (arc tan |). 10. Find the value of the following expressions : (a) sin (arc cos |). (c) cos (Arc cos y\). (e) sin (Arc sin \). (&) sin (arc sec 3). {d) sec (Arc esc 2). (/) tan (Arc tan 6). 11. Prove that Arc sin (2/6) = Arc tan (2/V2T). 12. Find x when Arc cos (2 x^ — 2 x) = 2 7r/3. Find the values of the following expressions : 13. cos [90^ — Arc tan |]. 14. sec [90° -Arc sec 2]. 15. tan [90° - Arc sin ^{\. 196 MATHEMATICAL ANALYSIS [VII, § 135 135. Projection. Consider two directed lines p and g in a plane, i.e. two lines on each of which one of the directions has been specified as positive (Fig. 123). Let A and B be any two points on p and let A\ B' be the points in which per- Fig. 123 pendiculars to q through A and B, respectively, meet q. The directed segment A'B' is called the projection of the directed seg- ment AB on q and is denoted by A'B' = proj^ AB. In both figures AB is positive. In the first figure A'B' is posi- tive, while in the second figure it is negative. As special cases of this definition we note the following : 1. If p and q are parallel and are directed in the same way, we have ^m]^AB=:AB. 2. If p and q are parallel and are directed oppositely, we have ^xo]^AB=-AB. 3. If p is perpendicular to q, we have proj,^B = 0. It should be noted carefully that these propositions are true no matter how A and B are situated on p. We may now prove the following important proposition; VII, § 135] TRIGONOMETRIC RELATIONS 197 If Aj B are any two points on a directed line p, and q is any directed line in the same plane with p, then we have both in Tnagnitvde and sign : (1) proj\ AB = AB' cos {qp), where (qp) represents an angle through ichich q must be rotated in order to make its direction coincide with the-direction of p. We note first that all possible determinations of the angle (qp) have the same cosine, since any two of these determina- tions differ by multiples of 360° (Fig. 124). We shall prove Fig. 124 the proposition first for the case where AB has the same direc- tion as p, i.e. where AB is positive. To this end we draw through A (Fig. 125) a line qi parallel to q and directed in the A' \Z. B' ^ A Bi ^ B J. I' J S' ' ^ Fig. 125 same way. (We may evidently assume without loss of gener- ality that q is horizontal and is directed to the right.) Let A'B' have the same significance as before and let BB' meet ^'i in Bi. Then, by the definition of the cosine, we have A 7? —^ = cos (q^p) = cos (qp)y in magnitude and in sign ; or ABi = ^5 cos (qp). But Therefore AB, = A'B'=VTOj^AB. projg AB = AB cos (qp). 198 MATHEMATICAL ANALYSIS [VII, § 135 Finally, if AB is negative, BA is positive, and, by the result just obtained, we should have B'A' == BA cos (qp). Hence, changing signs on both sides of this equation, we have A'B' = AB cos (q 2^). The special cases 1, 2, 3 listed on p. 196 are obtained from formula (1) by placing (qp) equal to 0°, 180°, 90°, respectively ; for cos 0° = 1, cos 180° - - 1, cos 90° = 0. 136. Application of Projection. In Physics, forces and velocities are usually represented by line segments. A force of 20 pounds, for example, is represented by a segment 20 units in length and drawn in the direction of the force. A velocity of 20 feet per second is represented by a segment 20 units in length and drawn in the direction of the motion. The projection on a given line I of a segment representing a force or velocity represents the component of the force or velocity in the direction of l. Example. A smooth block is sliding down a smooth incline which makes an angle of 30° with the horizontal. If the block weighs 10 lb., what force acting directly up the plane will keep the block at rest ? Draw the segment AB 10 units in length, directly downward to represent the force exerted by the weight. Project this segment Fig. 126 ^^ ^^^ incline and call this projection AC. Now angle ABC = 30°. Therefore AC = AB sin 30° = 5. This is the component of the force AB down the plane. Therefore a force of 5 lb. acting up the plane will keep the body at rest. A •^ B a! c" B' VII, § 136] TRIGONOMETRIC RELATIONS 199 Theorem. If A, B, C are any three points in a plarie, arid I is any directed line iii the plane, the algebraic sum of the projec- tions of the segments AB and BC on I is equal to the projection of the segment AC on I. As a point traces out the path, from A to B, and then from B to O (Fig. 127), the projection of the poiiit traces out the segments from A' to B' and then from B' to C. The net result of this motion is a motion from A' to C which represents the projection of AC, i.e. A'B' + B'C = A'C, Fig. 127 EXERCISES 1. What is the projection of a line segment upon a line I, if the line segment is perpendicular to the line I ? 2. Find proj^^ AB and proj^ AB* in each of the following cases, if a denotes the angle from the cc-axis to AB. (a) AB = 5, a = 60°. (c) AB = 6, a = 90°. (6) AB = 10, a = 300°. (d) AB = 20, a = 210°. 3. Prove hy means of projection that in a triangle ABC a = b cos C + c cos B. 4. If proja; AB = 3 and projy AB = — 4, find the length of AB. 6. A steamer is going northeast 20 miles per hour. How fast is it going north ? going east ? 6. A 20 lb. block is sliding down a 15° incline. Find what force acting directly up the plane will just hold the block, allowing one half a pound for friction. 7. Prove that if the sides of a polygon are projected in order upon any given line, the sum of these projections is zero. * Proji AB and projy AB mean the projections of AB on the x-axis and the y-axis, respectively. 200 MATHEMATICAL ANALYSIS [VII, § 137 137. Rotation in a Plane. Suppose that a point P{Xj y) in a plane moves on the arc of a circle with center at the origin 0, through an angle a. Suppose that its position after this rotation is Pix^ y') referred to the same axes of coordinates. We desire to find x' and y' in terms of a;, y, and a. In Fig. 128 we have drawn P and its coordi- nates X = OM, y = MP, and the new position OM' P' of the triangle OMP after a rotation about the origin through an angle a. The coordinates x' = ON, ?/' = NP' of P are the pro- jections of OP on the a;-axis and the y-axis re- spectively, and these pro- jections are equal respec- tively to the sum of the projections of OM' and M'P on the respective axes. Hence, x' = proj, OP = proj, OM' -f- proj, M'P = OM' cos {OX, OM') 4- M'P cos (OX, M'P) = X cos a -\- y cos (a -1- 7r/2) = X cos a — y sin a. y' = proj, OP = proj^ OM' -f- proj^ M'P' = OM' cos (OF, 03/') -f 3/'P' cos ( OY, M'P') = X cos (— 7r/2 + «) -f- 2/ cos a = X sin a -h y cos a. Therefore, if the point P{x, y) is rotated about the origin through an angle a, the coordinates (x', y') of its new position are given by the formulas Y> \ 1 P'l 1 ^ N. CC m k L / P hs\^ V '/>'''x' \ 1^ T X ¥ ^X Fig. 128 VII, § 138] TRIGONOMETRIC RELATIONS 201 (1) J x' = X cos a — y sin a [j/ = x sin a + 1/ cos a. It should be noted that the above method of derivation is entirely general, i.e. it will apply to a point P in any quad- rant and to any angle a. 138. The Addition Formulas. We may now enter upon a more detailed study of the properties of the trigonometric functions. We shall first express sin (a + p) and cos (a + /8) in terms of sin a, cos a, sin p, cos 13.* To this end let OP be the terminal side of any angle a (Fig. 129). If OP is then rotated about through an angle jS to the position OP, the terminal line of the angle a -\- p is OP'. If P has the coordinates (x, y) and P the coordinates {x\ y'), then from (1) § 137, x' = xcos p — y sin /8, y' = X sin ^ + y cos p. Now sin {a + y8) is by definition equal to ^ and cos (a 4- /?) OP' = OP. Hence sin(a+ fi)=^ = - sin/3 4-^ cos^S, 7- r r sin (a + P) = sin a cos p + cos a sin p. T / Mv) / '' p^ L^ V W" X X Fig. 129 to — where r r or (1) Also or (2) cos (a + ;8) = - = - cos ^ - ^ sin^, r r r cos (a 4- P) = cos a cos p — sin a sin p. ♦We have already had occasion to note that sin (a + ^) is not in general equal to sin a + sin ^. (See Ex. 5, p. 151.) 202 MATHEMATICAL ANALYSIS [VII, § 138 Further we have tan (a4- B) = sin (a -\- /?) __ sin « cos ^ + cos ce sin yg cos {a -f- y8) cos a cos /3 — sin a sin)8* Dividing numerator and denominator by cos a cos )8, we have (3) tan(a + 3) = ^^ + ^^ °^. Furthermore, by replacing yg by — /? in (1), (2), and (3), and recalling that sin (—/?)= — sin 13, cos (— ^)= cos /?, tan (— yS) = — tanyS, we obtain (4) sin(a— p)=sinacosp— cosasinp, (5) cos (a — p) = cos a cos p + sin a sin p, (6) tanCa- S)= tang-tanp v; i^ii^a p; i_^tanatanp EXERCISES Expand the following : 1. sin (45° + a)= 3. cos (60'' + a) = 5. sin (30° - 45°) = 2. tan (30°-^)= 4. tan (45° + 60°) = 6. cos ( 180° - 45°) = 7. What do the following formulas become if a = yS ? sin (a + /3) = sin a cos ^3 -f cos a sin /3. ^^^ /^ , on _ tan ct + tan j3 sin (a — /3) = sin a cos j3 — cos a sin jS. ' ~ 1 - tan a tan /3 * cos (a + /3) = cos a cos /3 - sin a sin ^. . , „. _ tan a — tan/3 lan fcc — p) — ■ ■ — — — — . cos (a — /3) = cos a cos ^ + sm a sin /3. 1 + tan a tan )3 8. Complete the following formulas : sin 2 a cos a + cos 2 a sin a = tan 2 « + tan a _ sin 3 a cos a — cos 3 a sin a = 1 — tan 2 a tan a ~ 9. Prove sin 75° =: ^^ + \ cos75°=^^-:\ tan75° = ^5+^. 2\/2 2V2 V3-1 10. Given tan a = |, sin /3 = ^\, and « and ^ both positive acute angles, find the value of tan (a -f /3) ; sin (a — /3) ; cos(a + /3) ; tan (a - /3). VII, § 138] TRIGONOMETRIC RELATIONS 203 11. Prove that (a) cos (60° + a) + sin (80° + «) = cos a. (b) sin (60° + 6)- sin (60° - ^) = sin d. (c) cos (30° -f ^) - cos (30° - d) = -^in d. (d) cos (45° + d)+ cos (45° -6) = V2 • cos 6. (e) sin(a + -J + sinfa — -j = sina. (/) cos(a + ^) + cos(a-^) = V3. (g) tan (45° + ^) = L±ia!L? . (h) tan (46° - 5) = ^ ^ ^^° ^ • 12. By using the functions of 60° and 30° find the value of sin 90° ; cos 90°. 13. Find in radical form the value of sin 15°; cos 15°; tan 16° J sin 105° ; cos 105° ; tan 105°. 14. If tan a = I, sin j3 = j\, and a is in the third quadrant while /3 is in the second, find sin (a ± j3) ; cos (a ± /3) ; tan (a ± /3). Prove the following identities : j^g sin (ct + /3) _ tan ct + tan /3 jg sin2cs ^^^ ^ ^ = ejn 3 « sin (a — j3) tan a — tan /3 ' sec a esc a j^ tan ot — tan (cc — /3) _ ^^ o 19- (a) sin (180° — &) = sin ^. l + tanatan(a -^) ~ (&) cos(180° - ^) =- cos^. 18. tan(0 ± 45°) + ctn {6 T 45°) = 0. (c) tan (180° - ^) = - tan Q. 20. cos (a + /3) cos (a — /3) = cos^ a — sin^ ^. 21. sin (a + jS) sin (a — /3) = sin2 a — sin2 /3. 22. ctn(«+^) = ^^"""^'^^-^ 23. ctn («-^) = ^ill^^^^+i . ctn a + ctn j3 ctn /3 - ctn a 24. Prove Arc tan ^ + Arc tan \ = ir/i [Hint : Let Arc tan l = x and Arc tan 1 = y. Then we wish to prove X -i-y = 7r/4, which is true since tan (x + y)= 1.] 25. Prove Arc sin a + Arc cos a = - , if < a < 1. 26. Prove Arc sin j\ + Arc sin f = Arc sin ||. 27. Prove Arc tan 2 -}- Arc tan ^ = 7r/2. 28. Prove Arc cos | + Arc cos (— ^j) = Arc cos ( — f f ). 29. Prove Arc tan j% + Arc tan | = Arc tan ||. 30. Find the value of sin [Arc sin | + Arc ctn |]. 31. Find the value of sin [Arc sin a -H Arc sin 6] if < a < 1, < 6 < 1. 204 MATHEMATICAL ANALYSIS [VII, § 138 32. Expand sin (x -{ y -\- z) ; cos(x + y + z). [Hint: a; + 2/ + 2 = (x + ?/) + 2.] 33. The area ^ of a triangle was computed from the formula A = I ab sin 6. If an error e was made in measuring the angle ^, show that the corrected area A' is given by the relation A' = A{co8€ + sinectn^). 139. Functions of Double Angles. In this and the follow- ing articles (§§ 139-141) we shall derive from the addition formulas a variety of other relations which are serviceable in transforming trigonometric expressions. Since the formulas for sin (a + /3) and cos (a + yS) are true for all angles a and fi, they will be true when JS = a. Putting /? = a, we obtain (1) sin 2 a = 2 sin a cos a, (2) cos 2 a = cos2 a — sin2 a. Since sin^ a + cos^ a = 1, we have also (3) cos 2 a = 1 - 2 sin2 a (4) = 2 C0S2 a — 1. Similarly the formula for tan (a + /?) (which is true for all angles a, ^, and a + /3 which have tangents) becomes, when )8 = «, ^ ^ 1 — tan^a which holds for every angle for which both members are defined. The above formulas should be learned in words. For ex- ample, formula (1) states that the sine of any angle equals twice the sine of half the angle times the cosine of half the angle. Thus sin 6 a; = 2 sin 3 x cos 3 a?, 2 tan 2 x tan 4 x = l~tan22a;' cos a; =5 cos*?— sin^?. VII, § 140] TRIGONOMETRIC RELATIONS 205 140. Functions of Half Angles. From (3), § 139, we have 2sin^^ = ; 1 — COS a. Therefore • «-!= (6) :±v'~r" From (4), § 139, we have Therefore 2cos2^ = cos| = 1 + COS a. (J) ±V'T"- rormulas (6) and (7) are at once seen to hold for all angles a. Now, if we divide formula (6) by formula (7), we obtain (8) ta"i=±Vr — cos a -j- cos a which is true for all angles a except n • 180°, where n is any odd integer. Example. Given sin ^ = — 3/5, cos A negative ; find sin (A/2). Since the angle A is in the third quadrant, A/2 is in the second or fourth quadrant, and hence sin (A/2) may be either positive or negative. Therefore, since cos A=— 4/5, we have 2 Al 2 VlO 10 EXERCISES Complete the following formulas and state whether they are true for all angles : 1. sin 2 a = 3. tan 2 a = 6. cos ^ = 2 2. cos2a= (three forms). 4. sin-i= 6. tan- = 2 ' ^ 7. In what quadrant is e/2 if 6 is positive, less than 360°, and in the second quadrant ? third quadrant ? fourth quadrant ? 206 MATHEMATICAL ANALYSIS [VII, § 140 8. Express cos 2 a in terms of cos 4 a. 9. Express sin x in terms of functions of 3 x. 10. Express tan 4 a in terms of tan 2 a. 11. Express tan 4 a in terms of cos 8 a. 12. Express sin x in terms of functions of x/2. 13. Explain why the formulas for sin x and cos x in terms of functions ot2x have a double sign. 14 From the functions of 30° find those of 60°. 15. From the functions of 60° find those of 30°. 16. From the functions of 30° find those of 16°. 17. From the functions of 15° find those of 7.5°. 18 Find the functions of 2 a if sin a = f and a is in the second quadrant. 19. Find the functions of a/2 if cos a =— 0.6 and a is in third quad- rant, positive, and less than 360°. 20. Express sin 3 a in terms of sin a. [Hint : 3a = 2a-f-a.] 21. From the value of cos 45° find the functions of 22.5". 22. Given sin a = — and a in the second quadrant. Find the values of 13 ^ (a) sin 2 a. (c) cos 2 a. (e) tan 2 a. (b) sin". (d) cos-. (/) tan-. 2 2 2 o 23. If tan 2 a = - find sin a, cos a, tan a if a is an angle in the third 4 quadrant. Prove the following identities : 24. L±-^2S^=ctn«. 27. 1 - cos 2 ^ + sin 2 g^ ^^^ ^ sin a 2 1 -I- cos 2 ^ 4- sin 2 ^ 26. fsin^-cos^j = 1 - sin 5. 28. sin^ + cos^ = ± Vl + sin«. 26. cos2g + cosg-H^^^^g 29. sec« + tan a = tanf? + «^ siu2e + sin^ \4 2/ 30. 2 Arc cos x = Arc cos (2 a;* — 1) . 31. 2 Arc coso; = Arcsin (2 arvl — a;2). VII, § 141] TRIGONOMETRIC RELATIONS 207 32. tan [2 Arc tan x] = -^-^ . 34. tan [2 Arc sec x] = ± ^ ^ 1-x-^ - 2 - x-^ 1 — r2 33. cos [2 Arc tan x] = ~ — • 35. cos (2 Arc sin a) = 1 - 2 a^. •■ -• 1 + x^ Solve the following equations : 36. cos 2 X + 5 sin X = 3. 40. sin^ 2 x — sin^ x = |. 37. cos 2 X - sin X = |. 41. sin 2 x = 2 cos x. 38. sin 2 X cos X = sin X. 42. 2 sin22x = 1 — cos2x. 39. 2 sin2 x + sin^ 2 x = 2. 43 ctn x — esc 2 x = 1. 44. A flagpole 50 ft. high stands on a tower 49 ft. high. At what dis- tance from the foot of tlie tower will the flagpole and the tower subtend equal angles ? 45. The dial of a town clock has a diameter of 10 ft. and its center is 100 ft. above the ground. At what distance from the foot of the tower will the dial be most plainly visible ? [The angle subtended by the dial must be as large as possible.] 141. Product Formulas. From § 138 we have sin (a 4- j8) = sin a cos ft -\- cos a sin p, sin (a — ^) = sin a cos ^ — cos a sin y8. Adding, we get (1) sin (a -f iS) + sin {a — 13)= 2 sin a cos jS. Subtracting, we have (2) sin (a -f /3) — sin (« — ^) = 2 cos a sin ft. Now, if we let a + /8 = P and a — fi = Q, then « = ^«, fi = ^. Therefore formulas (1) and (2) become . P 4- O P — O sm P H- sm Q = 2 sm — :r_J!^cos — — -^, 2 2 P 4- O . P— O Sin P — sm Q = 2 cos ^ ^ sm — — ^. ^ 2 2 Similarly, starting with cos (a + /3) and cos (a — ft) and per- forming the same operations, the following formulas result : 208 MATHEMATICAL ANALYSIS [VII, § 141 cos P + COS Q = 2 COS ^-±-2 COS ^^-=-2 2 2 COS P — COS p = - 2 sin "j" ^ sin ^ » 2 2 In words : the sum of two sines = twice sin (half sum) times cos. (half difference), the difference of two sines = twice cos (half sum) times sin (half difference),* the sum of two cosines = twice cos (half sum) times cos (half difference), the difference of two cosines = minus twice sin (half sum) times sin (half difference).* Example 1. Prove that cos 3^ +cosa;^^^^g sin 3 a; + sin x for all angles for which both members are defined. cos 3 a; + cosx _ 2 cos ^ (3 x + a:) cos \ (3 a; — a;) _ cos 2 a; _ ^ « ^ sin 3 X + sin x 2 sin \ (3 x + a:) cos ^ (3 x — x) ~ sin 2 x ~ Example 2. Reduce sin 4 x + cos 2 x to the form of a product. We may write this as sin 4 x + sin (90° — 2 x), which is equal to 2 ,i„ 4a;4-90°-2» ^„3 4»-90° + 2a; ^ ^ ^„ ^^^. + x) COS (3 « - 45°). 2 2d EXERCISES Reduce to a product : 1. sin 4 ^ — sin 2 ^. 4. cos 2 ^ + sin 2 6. 7. cos 3 x + sin 5 x. 2. cos + cos 3 d. 5. cos 3 ^ — cos 6 6. 8. sin 20° — sin 60°. 3. cos65 + cos2^. 6. sin (x 4- Ax) — sin X. Show that 9. sin 20° + sin 40° = cos 10°. ^^ sin 15° + sin 75° _ _ ^^ g^o 10. cos 50° + cos 70° = cos 10°. * sin 16° - sin 75° a. sin 75° -sin 15° ^^^3Qo^ 13 sin 3 g- sin5g^_ ^^^^^^ cos 75° + cos 15° cos 3 6 — cos 50 * The difference is taken, first angle minus the second. VII, § 142] TRIGONOMETRIC RELATIONS 209 Prove the following identities : 14. sin 4 ct -f- sin 3 ct _^^^ce jg sin « + sin /3 __ tan ^ (ce -H3) cos 3 a — cos 4 a 2 ", sina — sin/3~tau ^ (a — ^) 4 e cos a + 2 cos 3 a + cos 5 a cos 3 a JLO. ■ ■ = • cos 3 « + 2 cos 5 a + cos 7 a cos 5 a ^rj cosct-cos/3 _ tan|(« + /3) ^g sin (n - 2) g + sin w g _ ^^^ ^ cos a + cos ^ ctn i (a -|8) ' cos (n— 2)-0 - cosn^ Solve the following equations : 19. cos 6 4- cos 5 ^ = cos 3 ^. 22. sin 4 — sin 2 ^ = cos 3 d. 20. sin ^ + sin 5 ^ =: sin 3 d. 23. cos 7 ^ — cos ^ = — sin 4 6. 21. sin 3 ^ + sin 7 ^ = sin 6 6. 142. Law of Tangents. A method for shortening computa- tion will be presented in the next chapter. In applying this method to the solution of triangles the formulas given below are valuable. We shall state first the so-called law of tangents: The difference of two sides of" a triangle is to their sum as the tangent of half the difference of the opposite angles is to the tan- gent of half their sum. Proof. a^sin^, 5 sin 5 • Hence, by proportion, we have g — 6 _ sin ^ — sin ^ a 4- 6 sin ^ -f- sin 5 But . ^ . ^ 2cos^i±^sin^:^ tan^^ sm ^ - sm J5 2 2 2 sin ^-f sin ^ o - A -\- B A — B . A ~ 2 sm — cos tan — 2 2 tan^^ Therefore a-h^ 2_^ « + «> tan4+-§ D 210 MATHEMATICAL ANALYSIS [VII, § 143 143. Angles of a Triangle in Terms of the Sides. Con- ^ struct the inscribed circle of the triangle and denote its radius by r. If the perim- eter a + &4-c=2s, then (Fig. 130) AE = AF=s- a, .U-s-a^F B BD= BF= s-b. Fig. 130. CD = CE = s - c. Then tani^ = -^, tani.5 = -^^, taniC=^l-, s—a s — b ^ s—c where, from § 130, -■ J(^-«)(g-^)(^-c) ' MISCELLANEOUS EXERCISES 1. Reduce to radians 65°, — 135°, — 300°, 20°. 2. Reduce to degrees tt, 3 tt, — 2 tt, 4 tt radians. 3. Find sin (a — /3) and cos (a + ^3) when it is given that a and /3 are positive and acute and tan a = | and sec p = ^. 4. Find tan (a + /3) and tan (a — /3) when it is given that tan a = ^ and tan /3 = ^, 6. Prove that sin 4 a = 4 sin a cos a — 8 sin^ a cos a. 2 6. Given sin ^ = — ^, and d in the second quadrant. Find sin 2 d, V5 cos 2 ^, tan 2 ^. Prove the following identities : . 7. sin2a = -2iHL^. 8. cos2 « = ^ " ^^"'« 1 -f tan2 a 1 + tan'-^ a 9. sec2a = _5?^i«_ 10. tan«= "^"^" . csc2 a — 2 1 + cos 2 a 11. sin (a -\- j3) cos /3 — cos (a + p) sin /S = sin a. 12. sin 2 a + sin 2 )3 + sin 2 7 = 4 sin a sin /S sin 7, if a + j9 + 7 = 180°. 1 + tan - la co8« _ 2 1 — sin a I .. a VII, § 143] TRIGONOMETRIC RELATIONS 211 14. 1 + tan a tan ^ 15. 2 sin^ a + cos^ ct _ 2 — sin 2 ce sin a + cos a " 2 16. «-H?l^ = 2cos2«. sin 2 a 17. Arc cos f 4- Arc tan f = Arc tan W, Solve the following equations : 18. cos 2 a = cos^ a. 19. 2 sin a = sin 2 a. 20. cos 2 a + cos a = — 1. 21. sin a + sin 2 a + sin 3 a = 0. 22. sin 2 a — cos 2 a — sin a 4- cos a = 0. 23. Arc tan x + Arc tan (1 — x) = Arc tan |. 3 26. Arc tan ?-tl + Arc tan ^^ = 180° + Arc tan ( - 7). x — \ X 26. Arc sin x + Arc sin- = 120°. 2 2 tr 27. Arcsina; + 2 Arccosx = -^• 3 In a right triangle ABC, right angled at (7, prove 28. sin2 ^ = £jZi? . 29. cos^ ^ = .^±i: . 30. *tan ^ " 2 2c 2 2c 2 a + 6 31. Solve for x and ?/ the following equations : ic sin a + ^ cos a = sin a, X cos a — ?/ sin a = cos a. 32. Solve for x and y the following equations : a; cos ^ — y sin ^ = sin ^, X sin ^ + y cos ^ = cos d. 33. If 2 X is less than 90° and sinx=cos(2 x + 40°), find the value of x. 34. Find « so that the equation x^ + 2 x cos a + 1 = shall have equal roots, 35. Find a so that the equation 3 x^ + 2 x sec a + 1 = shall have equal roots. CHAPTER VIII THE LOGARITHMIC AND EXPONENTUL FUNCTIONS 144. The Invention of Logarithms. In the last two chap- ters we have had occasion to do a considerable amount of numerical computation. In spite of the fact that we have confined these computations to comparatively small numbers and have had the assistance of tables of squares and square roots, the calculations have often been laborious. To carry out by the methods thus far at our disposal the computations involved in many of the problems of insurance, engineering, astronomy, etc., would require a prohibitive amount of labor. That it is now practicable jjo effect such computations is largely due to the invention ©f logarithms by John Napier (1550-1617), Baron of Merchjfeton, in Scotland. As in the case of many epoch-making inventions, the funda- mental idea of Napier was extraordinarily simple. It may be explained as follows. Consider the function y = 2". We readily obtain the following table of corresponding values : (1) X 1 2 3 4 6 6 7 8 9 10 11 12 2/ = 2^ 2 4 8 16 82 64 128 256 512 1024 2048 4096 Now, since 2" • 2' = 2"+'', it is clear that, if we desire to ob- tain the product of two numbers in the lower line of the table, we need only add the two corresponding numbers in the upper line (the exponents), and then find the number in the lower 212 VIII, § 145] EXPONENTS — LOGARITHMS 213 line which corresponds to this sum* For example, to find the product of 128 x 16, we find from the table that the numbers corresponding to 128 and 16 are 7 and 4, respectively; the sum of the last pair is 11 and the number in the lower line corre- sponding to 11 is 2048, which is the product sought. Or again, to find 4096 -?- 512, we find the corresponding exponents 12 and 9 in the table, subtract (12 — 9 = 3), and find the required quo- tient to be 8. How would you justify the latter procedure ? While the fundamental idea here described is simple, con- siderable insight was required to make the idea practicable. For, the above table makes possible the finding of the product of two numbers only when the numbers in question and their product are to be found in the lower line of the table. In order to be useful in practical computation it is obviously necessary to construct a table which will contain every number, or at least from which the corresponding " exponent " of any number can easily be obtained either precisely or with a high degree of approximation. Th^ problem confronting Napier was to Jill in the gaps in the numbers of the lower line of the table on p. 212, while preserving the fundamental property of the table, yIz. that to the product of any two numbers of the lower line corresponds the sum of the two corresponding numbers of the upper line. 145. Extension of the Table. An examination of table (1) reveals the following properties : (a) the values of x form an arithmetic progression (A.P.), since every number after the first is obtained by adding 1 to the preceding number ; (6) the values of y form a geometric progression (G.P.), since every number after the first is obtained by multiplying the preceding number by 2. These considerations suggest the possibility of extending the table in two ways. 214 MATHEMATICAL ANALYSIS [VIII, § 145 -5 -4 -3 -2 - 1 • 1 2 o 4 5 6 1 7 0.03126 0.0625 0.125 0.25 0.5 1 2 4 8 16 32 64 128 In the first place, we may extend it to the left so as to make the lower line contain numbers less than 2. To do this, we need only subtract 1 successively from the numbers of the upper line and divide by 2 successively the numbers of the lower line. We then obtain a table extending in both directions : (2) This table is still satisfactory. If we desire to multiply 128 by 0.0625, we add the corresponding numbers of the upper line, namely, 7 and — 4 ; thus we obtain the number 3, which according to the previous rule should give 128 x 0.0G25 = 8, which is correct. That the rule still applies may be tested on other products ; the fact that it does will be proved later. In the second place we may find new numbers to fill the gaps in the original table, by inserting arithmetic means between the successive values of x and geometric means between the successive values of y. Thus, if we take the following portion of the preceding table -2 -1 1 2 3 4 i i 1 2 1 4 8 IG and insert between every two successive numbers of the upper line their arithmetic, and between every two successive num- bers of the lower line their geometric mean, we obtain the table (3) -2 -f - 1 -h i 1 3 2 5 -2 3 1 4 i ^\/2 h iV2 1 V2 2 2V2 4 4V2 8 8V2 16 VIII, § 145] EXPONENTS -— LOGARITHMS 215 If the radicals are expressed approximately as decimals, this table takes tlie form -2.0 -1.5 - 1.0 -0.5 ' 0.5 1.0 1.5 2 2.5 3 3.5 4 0.25 0.35 0.50 0.72 l.OC 1.41 2.00 2.8S 4.00 5.66 8.00 11.31 16 x(AP.) 0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00 2.25 yCG.R) 1.00 1.19 1.41 ' 1.68 2.00 2.38 2.83 3.36 4.00 ' 4.76 1 Repeating this process of inserting means, we get the follow- ing table. To save space, we have begun the arithmetic pro- gression with and the geometric progression with 1, and have not carried the table as far as in the preceding case. (4) The rule for multiplying two values of y seems to apply also to this table, at least approximately. For example, if we apply the rule to find 3.36 x 1.19, we note that the sum of the cor- responding values of x is 1.75 -f 0.25 = 2.00 and conclude that 3.36 X 1.19 = 4.00. Actual multiplication gives 3.36 x 1.19 = 3.9984. The discrepancy we may attribute to the fact that the values of y other than 1, 2, 4 are only approximations to the true values.* The process used in constructing this table may be continued indefinitely. It enables us to interpolate a new value of x be- tween any two successive values of x and a new value of y between the two corresponding values of y. But this means that we can make the values of x and y as dense as we please, in other words, we can make the difference between successive values of y as small as we please. By continuing the process ♦ In fact the rules for computing with approximate numbers would lead us to write 4.00 in place of 3.9984 as we have no right to retain more than two decimal places. See § 160. 216 MATHEMATICAL ANALYSIS [VIII, § 145 long enough we can make any number appear among the values of y to as high a degree of approximation as we desire and our intention of filling the gaps will then be attained. We must now prove, however, that the rule for multiplication does really hold in the extended table. Thus far we have merely verified this rule for special cases. EXERCISES 1. Assuming that the rule for multiplication applies, find by means of table (4) the following products. 3.36 X 1.41, 1.68 X 2.38, (1.68)2, (1.19)6. Check by ordinary multiplication. 146. Arithmetic and Geometric Progressions. The tables constructed consist of an arithmetic progression one term of which is the number (the terms of this arithmetic progression we denoted by x) and a geometric progression one term of which is the number 1 (the terms of this geometric progression we denoted by y). Moreover, to every value of x corresponds a definite value of y in such a way that to oj = corresponds 2/ = 1, and that to each succeeding (or preceding) value of x corre- sponds the succeeding (or preceding) value of y. Now suppose that the common difference of the arithmetic progression is d and that the common ratio of the geometric progression is r.- The correspondence between the values of x and y would then be exhibited in the following table. (5) We shall now prove that in this tahle^ to the product of any two valiLes of y corresponds the sum of the two corresponding values ofx. X ... -md ... -3d -2d -d d 2d Sd ... nd ... y ... 1 ... 1 1 1 r 1 r r^ r8 r" ... Vni, § 147] EXPONENTS — LOGARITHMS 217 If tlie two values of y are both, to the right of y = 1, for ex- ample i/i = r^, y.2 = J*^, then the corresponding values of x are pd and qd. To the product 2/12/2= ^^^'^ corresponds {p-\-q)d. If the two values of y are both to the left of 2/ = 1, the proof is similar. It is left as an exercise. If one value is to the left of 2/ = 1, for example, y — l/r^, and the other value is to the right, for example 2/2 = ''^j the cor- responding values of x are — pd and qd respectively. The product 2/12/2 is equal to (1/r^) r« = r^~^ if q> p, and is equal to l/r^« if q<p. The value of x corresponding to 2/12/2 is then {q — p)d/\i q> p and —{p — q)d ii q<,p. But {q — p)d = — {p —q)d = qd -\-{— pd). The discussion of the case p = q\^ left as an exercise. If one of the values of y is 1, the desired result follows immediately. Why ? In view of this theorem the validity of the rule used in the last article for multiplication is established. For tables (2) and (4) are both tables of the type (5), the former having d = 1 and r = 2, the latter having d — 0.25 and r = V2 = 1.19 (approximately) . 147. The Exponential Function (i^{a > 0). Let us now con- sider the table X — m ... -3 -2 - 1 1 2 3 ... n ... y ... 1 a3 1 a2 1 a 1 a a2 a^ a'» ... where a represents any positive number.* This table defines y as a function of x. Morover, this table is a table of the type (5) ; and all tables obtained by interpolating arithmetic means between two successive values of x and the same number of geometric means between the corresponding values of y are of * The value a = 1 leads to trivial results. Hence, we assume also that a ^ 1. 218 MATHEMATICAL ANALYSIS [VIII, § 147 the type (5). Thus if we interpolate q arithmetic means be- tween x — and x = 1, and q geometric means between y —1 and y= a, we obtain the following table : X ... ... -1 -2 _1 1 Q 2 1 ... y ... 1 ... 1 a 1 1 1 la ^.'/;;^2 ... V^f i 1 X 1 9 + 1 Q ... 2 ... . iv'ar' a {Vay^' ... a2 {VaY ... which is a table of the type (5), with d = 1/q and r = va. The function y oi x thus defined is y = a% for ic = 1, 2, 3, •••. We are therefore led to define the expression a* for fractional and negative values of x and for x = as follows : (1) aO = l. (2) ai/« means ^o", where q^ is a positive integer. (3) aP/^ means (Va)^, or its equal V^y where p and q are positive integers. (4) a-^ means 1/a**, where n is any positive rational number. In view of the fundamental property of any table of type (5), whereby to the product of any two values of y corresponds the sum of the two corresponding values of x, we have (t^ . av — flM+f for^all values of u and v for which the expressions a", a", and a""*"' have been defined. The function ?/ = a* (a > 0) has now been defined for all rational values of x. To complete the definition of this func- * We should keep in mind that the symbol ^a (o>0) means the positive gth root of a. Thus yfm = 2, not — 2. VIII, §147] EXPONENTS — LOGARITHMS 219 tion for all real values of a;, we must indicate the mean- ing of a* when x is an irrational number. To carry this definition through in all its details is beyond the scope of an elementary course. But we have seen that any irrational number may be represented approximately by a rational num- ber, with an error as small as we please. (See § 29.) Thus V3 is represented approximately by the rational numbers 1.7, 1.73, 1.732, ..-. Our previous definitions have given a definite meaning, for example, to 2^-^, 2^-''^, 2'^-''^^, .... The values of the latter expressions are by definition approximate values of 2^^. We take for granted without proof the fact that the successive numbers (6) 2^'\ 21-7^ 21-"^, ..., as the exponents represent closer and closer approxima- tions to V3, approach closer and closer to a definite number. This definite number is by definition the value of 2^. Similar considerations apply to the definition of a^, where a is any positive number and x is any irrational number. The principle involved is briefly expressed as follows : An approximate value of x gives an approximate value of a'. The value of a* can he found as accurately as we please by using a sufficiently accurate approximation to x. The objection might be raised that the calculation of '2?''^ involves the extraction of the 10th root of 2 and the calculation of 2^-'^ involves the extraction of the 100th root of 2, etc., and perhaps we do not know how to extract these roots. As a matter of fact we can calculate 2^^ as ac- curately as we please by extracting square roots only. The processus as follows : We know that \/3 = 1.7320 accurately to four decimal places. Now by table (4), p. 215, we see that 2^-^-2M and 2'-" = 3.36. We carry the computation to more places and have 2i«)oo =2.8284 and 2i-7500= 3.3636. Now, 1.7320 lies between 1.5000 and 1.7500, the arithmetic mean of which is 1.6250. The geometric mean of 2.8384 and 3 3635 is 3.0844 According to our previous definitions we have then 2i-<'25o = 3.0844. 220 MATHEMATICAL ANALYSIS [VIH, § 147 Inserting means between the last two results we have 2^-^^ = 3.2200. By inserting arithmetic means between the properly selected exponents and geometric means between the corresponding powers of 2 we can ulti- mately obtain the value of 2i-^320, ^he results of the necessary steps are : 21.7188 :^ 3.2915, 21-7344 = 3.3274, 21-7266 = 3.3094, 2i-7305 = 3.3182, 21.7325 == 3.3228, 21-7315 = 3.3205, 2i-7320 = 3. 32 17. The process here illustrated makes it possible to calculate 2^3 to as high a degree of approximation as we please, since we can carry the computa- tion to as large a number of decimal places as we please. 148. The Laws of Exponents. The function y = a'' (a > 0) is now defined for all real values of x. This function is called the exponential function of base a. The laws of exponents I. a^ ' a^ = a^^^ ] II. (cr*)'' = a«*» , a > 0, 6 > 0, III. a«* . b^ = (ab)^ ' which were derived previously (§ 42) for positive integral ex- ponents, hold for all real values of u and v. The first of these we have already derived. The last two may be readily proved for negative, fractional, and zero exponents by using the defini- tion of a''. Thus by definition, if it = p/q and v = 71, where p, q, n are positive integers, we have p pn If u is any positive rational number and v=p/qf where p, q are positive integers, we have, up (a^y = (a«)« = -Via'^y = -^/a"^ = a «" = a"". If u = — n, where n is a positive rational number, and if V =p/q, where p and q are positive integers, we have (a-y = (a-«)^/'=/^\/iY= — i— = -i = a" 7 = a«^ If u is any rational number and v =— n, where n is any VIII, § 149] EXPONENTS — LOGARITHMS positive rational number, 221 (a«)''=(a'*)-" — — — = — = a""" = a* If either w or v is zero, the result is immediate. Hence the law II is proved for rational exponents. A similar proof of the law III is left as an exercise. 149. The Graph of the Exponential Function. Figure 131 represents the graph of the function y = 2*, drawn from the tabular representation given in the first table on p. 215. ~ - 7 r ^ ~ n \ l. 1 \ u \ [e ri//^/^ h -12 \ R - «l \\ .U 's i \ i / \ / / ' / / ?^ / y j / "^ / 1 / r \ ' / / / \\ / / / / <^ \ A \ / y <\ / ' 1 • '5' N s^ V. / X "*" 1 y ^ ^ ^ a = 1 ^ 1 ^ ^ ?J , _ —J -4 ' V =2tc V = ^ _ Fig. 131 FiQ. 132 It will be noted that all curves of the system ?/ = a* pass through the point (0, 1). By hypothesis a > 0. If a > 1, the function a^ is an increasing function ; while if a < 1, the func- tion is a decreasing function. Figure 132 shows some of the curves of the system y = a*. 222 MATHEMATICAL ANALYSIS [VIII, § 149 EXERCISES 1. Calculate the value of the exponential function 3* (a) for the values of a; = 1, 2, 3, 4, 0, — 1, — 2, - 3, — 4 ; (b) for the values x = 0.5, 1.5, 2.5, 3.5, - 0.5, - 1.5, - 2.5 ; (c) for the values x = 0.25, 0.75, 1.25, 1.75. Arrange the results in the form of a single table. 2. Show how to use the table constructed in Ex. 1 to solve problems in multiplication, division, raising to powers, and extracting roots. Make up your own problems and check your results by the methods of arithmetic. 3. Describe in detail how you would find the value of 3 . Between what two numbers in the table found in Ex. 1 does the value of 3^^ lie ? 4. Construct the graph of y = 3* for values of x between — 2 and 3. 6. What is meant by a^ ? x^ ? (l/y)^ ? 6. What is the value of 8^ ? 27^ ? (0.001)^ ? (i)8 ? 7. Simplify (18)^ -i- (3)^' 8. Perform the following indicated operations : («) (^¥. w (^>^-K \x!^y~^y (6)(aWc¥. (,)^^_,+ 2,.i).. (c) (32x02/10)^. (/) (2^)*. 9. Multiply (a) {a-^ + a)ia-^-a). (6) (a-i-ao)(a-2-a'0(a-»-aO- (c) (a;^-y^)(a;« -y^). (d) (x-i 4- x~^y~^ + jr^Xx-i - x~^y~^ + y-^). 8 2 12 1 (e) (a* - 2 a* + 3 a*) (2 a'- a* + 2). 8 2 12 1 (/) (ir - oir + 3 6y« - c) (jr + 6y« - cyO). 10. Divide (a) (a;+l)by (v^ + 1). (&) (aj^' - y*) by (x^- yi^). (c) (J - a6^ + ah - b^) by (a^ - 6^). (d) (a-i + 4 a^ + 6 a^ + 4 a^ + a*) by (a"^ + a). VIII, § 150] EXPONENTS — LOGARITHMS 223 11. Simplify (a) 12" + I - 9-' + ^ + 27*. (6) ( '"'P ]'*. ^64 m-^p^ 12. Simplify 13. Which of the two numbers V5 and ^/8 is the greater and why ? 14. Simplify (2^ x2^)-f-(54)i 15. Prove that, if _ 1 Tx^ _ x"^"| 2Lj „-d' ■y y then 2vxy 16. Reduce to simplest form (c) (a^ + X*) (a2 _ x^)~^ - (a2 _ x^)^. 150. Definition of the Logarithm. The logarithm of a number JV to a base 6 (6 > 0, ^^t: 1) is the exponent x of the power to which the base h must be raised to produce the number N. That is, if then X — logbN. These two equations are of the highest importance in all work concerning logarithms. One should keep in mind the fact that if either of them is given, the other may always be inferred. 224 MATHEMATICAL ANALYSIS [VIII § 150 Tlie graph of the logarithm function (Fig. 133) is obtained from the graph of the corresponding exponential function by simply turning the latter graph over about the line through the origin bisecting the first and third quadrants. Y - , <- .— ■ * ^ ^ ^ x" '" / ^ / f 1 J 1 > 1 s ^^x\ ■ 1 V = z ff^ X . _ _ _J _ Fig. 133 EXERCISES 1. When 3 is the base what are the logarithms of 9, 27, 3, 1, 81, ^, 2. Why cannot 1 be used as the base of a system of logarithms ? 3. When 10 is the base what are the logarithms of 1, 10, 100, 1000 ? 4. Find the values of x which will satisfy each of the following equalities : (a) logs 27 = X. (d) loga a = x. (g) logs x = 6. (6) log^3 = 1. (e) logal = x. {h) logssx = i (c) log, 6 = i. (/) logaij^r = «• (0 log.ooi x = .00001. 6, Find the value of each of the following expressions : (a) loga 16. (c) logesi-g. (e) log26l26. (6) log34a49. (d)log2Vl6. (/) log2^. VIII, § 151] EXPONENTS — LOGARITHMS 225 151. The Three Fundamental Laws of Logarithms. From the properties of tlie exponential function (p. 220) we derive the following fundamental laws. I. Tlie logarithm of a product equals the sum of the logarithms of its factors. Symbolically, lege, MN = logft Af + logft N. Proof. Let logj, M= x, then 6* = M. Let log^ N~y, then b"=N. Hence we have MN= ft*"^", or logj,i¥iV=a; + 2/, i.e. \o^^,MN=\og^,M+\o^^N. II. The logarithm of a quotient equals the logarithm of the dividend minus the logarithm of the divisor. Symbolically M ^^Sft TT = logft M — logft N. N Proof. L^t logj, Jf = x, then b' = M. Let logj, -^ = y, then h" = N. Hence we have M/N = &*"»', or M M ^og,—=x-y, i.e. log, — = log,Jf-log,JV IIL T%e logarithm of the pth power of a number equals p times the logarithm of the number. Symbolically log6MP = />log5M. Proof. Let log^ M^x, then 6^ = M. Raising both sides to the pih. power, we have b^ = M^. Therefore logj, M^ =px=p log, M. From law III it follows that the logarithm of the real positive nth root of a number is one nth of the logarithm of the number. Q 226 MATHEMATICAL ANALYSIS [VIII, § 151 EXERCISES 1. Given logio 2 = 0.3010, logio 3 = 0.4771, logio 7 = 0.8451, find the value of each of the following expressions: (a) logio 6. (/) logio 6. [Hint: logio 2 X 3=logio 2+logio3.] [Hint : logio5 = logio V-] (6) logio 21.0. (.9) logio 150. (c) logio 20.0. W logio Vli. {d) logio 0.03. (i) logio 49. («) logio 1. U) logio V2V7^. 2. Given the same three logarithms as in Ex. 1, find the value of each of the following expressions: f^\ i«„ 4 x5 x7 ., . i^„ 5 X 3 X 20 ,. , ^ 2058 {d) logio (2)25. (e) logio (3)8(5)6. (/) logio (28)(|). 152. The Systems most Frequently Used. From the defi- nition of a logarithm (§ 150) any positive number except 1 can be used as the base of a system of logarithms. As a matter of fact, however, the numbers generally used are (1) a certain irrational number which is approximately equal to 2.71828 and is denoted by e and (2) the number 10. Logarithms to the base 6 are important in certain theoretical problems ; loga- rithms to this base are called natural. For numerical compu- tation it will be seen presently that the base 10 has numerous advantages. Since different systems of logarithms are in use, it is important to know how to change from one system to another. The following law explains how this can be done. IV. The logarithm of a number M to the base b is equal to the logarithm of M to any base a, divided by the logarithm of b to the ba^e a. Symbolically, lOga O VIII, § 153] EXPONENTS — LOGARITHMS 227 Proof. Let logj, M=x, then ¥ — M. Taking the logarithms of both sides to the base a, we have log^ ?>=^ = log„ 3f, or a; log„ 6 = log, 3f, log« h 153. Logarithms to the Base 10. Logarithms to the base 10 are known as common logarithms, or as Briggian logarithms, after Henry Briggs (1556-1631) who called attention to the advantages of 10 as a base. These advantages appear below. If 10 is the base, log 10 = 1, log 100 = 2, log 1000 = 3, etc. It follows that if a number be multiplied by 10, or by any positive integral power of 10, the logarithm of the number is increased by an integer. In other words, the shifting to the right of the decimal point in a number changes only the in- tegral part of the logarithm and leaves unchanged the decimal part of the logarithm. An example will make this clear. Given logio 2 = 0.30103, we have logio 20 = 1.30103, logio200 = 2.30103, etc. Or, again, given logio 4.5607 = 0.65903, we have logio 45.607 = 1.65903, logio 456.07 = 2.65903, logio 4560.7 = 3.65903, logio 45607.0 = 4.65903. It should be clear from these examples that the decimal part of the logarithm of a number greater than 1 in this system depends only on the succession of figures composing the num- ber, irrespective of where the decimal point is located ; while the integral part of the logarithm of the number depends simply on the position of the decimal point. The decimal part of a logarithm is called its mantissOj the integral part its characteristic. In view of what has been said above only the mantissas of logarithms to the base 10 need be tabulated. The characteristic can be found by inspection. This follows from the following considerations. 228 MATHEMATICAL ANALYSIS [VIII, § 153 The common logarithm of a number between 1 and 10 lies between and 1. The common logarithm of a number between 10 and 100 lies between 1 and 2. The common logarithm of a number between 100 and 1000 lies between 2 and 3. The common logarithm of a number between 10" and lO'*'^^ lies between n and n + 1. It follows that a number with one digit (^0) at the left of the decimal point has for its logarithm a number equal to + a decimal ; a number with two digits at the left of its decimal point has for its logarithm a number equal to 1 + a decimal ; a number with three digits at the left of the decimal point has for its logarithm a number equal to2-f- a decimal, etc. We conclude, therefore, that the characteristic of the common loga- rithm of a number greater than 1 is one less than the number of digits at the left of the decimal point. Thus, as before, logio 456.07 = 2.65903. The case of a logarithm of a number less than 1 requires special consideration. Taking the numerical example first con- sidered above, if logjo 2 =0.30103, we have logio 0.2 = 0.30103-1. Why ? This is a negative number, as it should be (since the logarithms of numbers less than 1 are all negative, if the base is greater than 1). But, if we were to carry out this subtraction and write logjo 0.2 = — .69897 (which would be correct)^ it would change the mantissa, which is inconvenient. Hence it is customary to write such a logarithm in the form 9.30103 - 10. If there are n ciphers immediately following the decimal point in a number less than 1, the characteristic is — n — 1. For convenience J if n < 10, we write this as (9 — n) — 10. This Vni, § 154] EXPONENTS — LOGARITHMS 229 characteristic is written in two parts. The first part 9 — n is written at the left of the mantissa and the — 10 at the right. In tlie sequel, unless the contrary is specifically stated we shall assume that all logarithms are to the base 10. We may accordingly omit writing the base in the symbol log when there is no danger of confusion. Thus, the equation log 2 = 0.30103 means logjo 2 = 0.30103. 154. Use of Tables. Since the characteristic of the loga- rithm of a number may be found by inspection, a table of logarithms contains only the mantissas. To make practical use of logarithms in computation it is necessary to have a con- veniently arranged table from which we can find (a) the logarithm of any given number, and (6) the number corre- sponding to a given logarithm. Tables of logarithms differ according to the number of decimal places to which the man- tissas are given and also in incidental details. However, the general principles governing their use are the same. These principles are explained for a four-place table (p. 536) by the following examples. Problem A. To find the logarithm of a given number, (1) When the number contains three or fewer figures. Example. To find the logarithm of 42.7. First, by §153, the characteristic is 1. We accordingly write (provi- ^'°°^'y^ log42.7 = l. Next we look up in the tables the mantissa corresponding to the succes- sion of figures 4, 2, 7. We run a finger down the first column of the table until we reach the figures 4, 2, hold it there while with another finger we mark the column headed with the third figure, 7. At the intersection of the line and column thus marked, we find the desired mantissa : 6304, The desired result is then log 42.7 = 1.6304. 230 MATHEMATICAL ANALYSIS [VIII, § 154 To find the logarithm of 0.0427, we should proceed in precisely the same way, the only difference being that the characteristic is now 8 — 10. H®^<^®' log 0.0427 = 8.6304 - 10. (2) When the number contains four significant figures Example. To find log 32.73. We see that again the characteristic is 1, and we write provisionally log32.73 = l. Now, the mantissa of log 32.73 lies between the mantissas of log 32. 70 and log 32.80; i.e. (from the table) between 5145 and 5159. The difference between these two mantissas (called the tabular difference at that place in the table) is 14, und this difference corresponds to a difference in the numbers of .10. According to the principle of linear interpolation,* the difference in the mantissas corresponding to a difference in the numbers of .03 is 14 X .3 = 4.2 or (rounded) 4. The mantissa corresponding to 3273 is then 5145 + 4 = 6149, and we obtain log 32.73 = 1.5149. Problem B. To find the number corresponding to a given logarithm. Here we simply reverse the preceding process. Example. To find the number whose logarithm is 0.8485. We first seek the mantissa 8485 in the table. We find that it lies be- tween 8482 and 8488, corresponding respectively to the successions of figures 7050 and 7060. The tabular difference here is 6, while our differ- ence, i.e. the difference we have to account for (8485 — 8482) is 3. Hence the corresponding difference in the numbers is | of 10 or 6. Hence the succession of figures in the number sought is 7055. Since the char- acteristic is 0, the number sought is 7.055. Or, log 7.055 = 0.8485. If the mantissa is found exactly in the table, of course no interpolation is necessary. Thus the number whose logarithm is 9.7348 — 10 is 0.5430. EXERCISES 1. Find the logarithms of the following numbers from the table on pp. 536-7 : 482, 26.4, 6.857, 9001, 0.5932, 0.08628, 0.00038. 2. Find the numbers corresponding to the following logarithms : 2.7935, 0.3502, 7.9699 - 10, 9.5300- 10, 3.6698, 1.0958. * One should convince oneself that the conditions for linear interpolatiou are satisfied by this table. In fact, it is readily seen that for several numbers immediately preceding and following 327, the tabular differences are 13 and 14. VIII, § 155] EXPONENTS — LOGARITHMS 231 155. Use of Logarithms in Computation. The way in which logarithms may be used in computation will be sufficiently ex- plained in the following examples. A few devices often neces- sary or at least desirable will be introduced. The latter are usually self-explanatory. Reference is made to them here, in order that one may be sure to note them when they arise. The use of logarithms in computation depends, of course, on the fundamental properties derived in § 151. Example 1. Find the value of 73.26 x 8.914 x 0.9214. We find the logarithms of the factors, add them, and then find the number corresponding to this logarithm. The work may be arranged as follows : Numbers Logarithms 73.26 (->) 1.8649 8.914 (-» 0.9501 0.9214 (-» 9.9645 - 10 12.7795-10 601.9 ^ws. ' (^) 2.7795 Product Example 2. Find the value of 732.6 ^ 89.14. Numbers Logarithms 732.6 (^^) 2.8649 89.14 (->) 1.9501 Quotient = 8.219 Arts. (<-) 0.9148 Example 3. Find the value of 89.14 -- 732.6. Numbers Logarithms 89.14 (->>) 11.9601 - 10 732.6 (-^) 2.8649 Quotient = 0.1217 Ans, (<~) 9.0852 - 10 Example 4. Find the value of 763.2 x 21.63 Whenever an example involves several different operations on the logarithms as in this case, it is desirable to make out a blank form. When a blank form is used, all logarithms should be looked up first and entered in their proper places. After this has been done, the necessary opera- tions (addition, subtraction, etc.) are performed. Such a procedure saves time and minimizes the chance of error. 232 MATHEMATICAL ANALYSIS [VIII, § 155 Form Numbers Logarithms 763.2 (-» 21.63 (-» (+)..... product 986.7 Ans. (-» (-) ." '.'.'. .... (<-) Form Filled In Numbers Logarithms 763.2 (->) 2.8826 21.63 (->) 1.3351 product 4.2177 986.7 (->) 2.9942 16.73 Ans. i<-) 1.2235 Example 5. Find (1.357)6. Numbers Logarithms 1.357 (->) 0.1326 (1.357)6 : = 4.602 Ans. (-^) 0.6630 Example 6. Find the cube root of 30.11. Numbers Logarithms 30.11 (_>) 1.4787 y/SOAl = 3.111 Ans. «-) 0.4929 Example 7. Find the cube root of 0.08244. Numbers Logarithms 0.08244 (->) 28.9161-30 v^O.08244 = 0.4352 Ans. (-^) 9.6387 - 10 EXERCISES Compute the value of each of the following expressions using the table on pp. 536-537. 1. 34.96 X 4.65. 5. (34.16 x .238)2. 2. 518.7 X 9.02 x .0472. 6. 8.572 x 1.973 x (.8723)8. „ 0.5683 0.3216 6.007 X 2.483 6.524 X LUO' 7. 648.8 ^(21.4)2 \2791 Vm, § 155] EXPONENTS — LOGARITHMS 233 9. 10. 11. J 2.8076X 3.184 ^ (2.012)3 »/ 2941 X 17^32 '>'2173 X 18.76* \/0. 00732 V735 ^^ 12, (20.027)i. (2.01)i 17. The stretch s of a brass wire when a weight m is hung at its free end is given by the formula , where m is the weight applied in grams, g = 980, I is the length of the wire in centimeters, r is the radius of the wire in centimeters, and yfc is a constant. If m = 844.9 grams, I — 200.9 centimeters, r = 0.30 centi- meters when s = 0.056, find k. 18. The crushing weight P in pounds of a wrought iron column is given by the formula -^ 55 P = 299,600^^, where d is the diameter in inches and I is the length in feet. What weight will crush a wrought iron column 10 feet long and 2.7 inches in diameter ? 19. The number n of vibrations per second made by a stretched string is given by the relation . , — 2lM m' where I is the length of the string in centimeters, ilf is the weight in grams that stretches the string, m the weight in grams of one centimeter of the string, and g =980. Find n when i{f= 5467.9 grams, Z = 78.5 centi- meters, m = 0.0065 grams. 20. The time t of oscillation of a pendulum of length I centimeters is given by the formula t = rJ-L. >'980 Find the time of oscillation of a pendulum 73.27 centimeters in length. 21. The weight w in grams of a cubic meter of aqueous vapor saturated at 17° C. is given by the formula ^ ^ 1293 X 12.7 X 5 computer. (1+M)(760x8)- 234 MATHEMATICAL ANALYSIS [VIII, § 156 156. Exponential Equations. An equation in which the unknown is contained in an exponent is known as an exponen- tial equation. Some such equations may be solved by taking the logarithms of both sides after the equation has been properly transformed. 2x4-1 Example 1. Solve the equation 3 + 7 = 15. Transposing the 7 and taking logarithms of both sides we obtain {2x + l)log3 = log8. Hence we find ^^iH-^^ij. 2Llog3 Example 2. Money is placed at interest, compounded annually. Find a formula for the amount at the end of n years. Also a formula* giving the number of years necessary to produce a given amount. Let C be the original capital and r the given rate of interest {i.e. if the interest is 5 per cent, r = 0.05). The amount A\ at the end of the first year is ^ ^ ^ ^,, Ai= G+ Cr= C{\ + r). At the end of two years we have ^2 = ^1(1 + = 0(1 + 02. At the end of n years, the amount is A=A,= G{\^rY. This is the formula required. To find w, given A, O, r, we take the logarithms of both sides and find EXERCISES 1. Solve for x the equation 2^ = 5. 2. Solve for y the equation 3w + 2 = 9. ■ 3. Solve for x and y the simultaneous equations S^+v = 4, 2*~v = 3. 4. Solve for x and y the simultaneous equations 2*+f = 6y, 3*-^ = 2^+^ 6. Find the amount of $1000 in 25 years at 6 per cent compound interest, compounded annually. VIII, § 1 56j EXPONENTS — LOGARITHMS - 235 ^ 6. Find the amount of $ 600 in 10 years at 4 per cent compound inter- est, compounded semiannually. 7. In how many years will a sum of money double itself at 5 per cent interest compounded annually ? semiannually? 8, A thermometer bulb at a temperature of 20° C. is exposed to the air for 15 seconds, in which time the temperature drops. 4 degrees. If the law of cooling is given by the formula 6 = ^oe~^S where is the final tem- perature, do the initial temperature, e the natural base of logarithms, and t the time in seconds, find the value of b. MISCELLANEOUS EXERCISES 1. "What objections are there to the use of a negative number as the base of a system of logaiithms ? 2. Show that a^^g^"^ = x. 3. Write each of the following expressions as a single term : (a) log x-\-\ogy — log z. (6) 3 log a; — 2 log y + 3 log z. (c) 3 log a - log (x + y)- I log (ex + d)+ log VwT~x. 4. Solve for x the following equations : (a) 2 log2 X -h log2 4 = 1. (c) 2 logio x — 3 logio 2 = 4. (b) logs « - 3 logs 2=4. {d) 3 logs x + 2 loga 3=1. 5. How many digits are there in 235 ? 3142 ? 312 ^28? 6. Which is the greater, (il)^^^ or 100 ? 7. Find the value of each of the following expressions. (See § 152. ) (a) logeSS. (6) logs 34. (c) log7 246. (cZ) logi3 26. 8. Prove that logb a • logo b = 1. 9. Prove that log, x-{-Vx^-l ^ 2 log, lx+ v^^^l]. X — y/x'^ — 1 10. The velocity v in feet per second of a body that has fallen s feet is given by the formula v = \/64.3s. What is the velocity acquired by the body if it falls 45 ft. 7 in. ? 11. Solve for x and y the equations : 2^ = IGv, aj + 4 y = 4. CHAPTER IX NUMERICAL COMPUTATION I. ERRORS IN COMPUTATION 157. Absolute and Relative Errors. In § 29 we noted that the numerical result of every observed measurement is an approximation. The difference between the exact value of the magnitude and this observed value is a concrete number called the absolute error* Often the absolute error is not the most serviceable measure of the precision of a measurement. The relative error, which is defined as the ratio of the absolute error to the exact value, is often found more serviceable. Since the relative error is a ratio, it is an abstract number, and is therefore sometimes expressed in per cent. For example, if the diagonal of a square 10 in. on a side be measured and and found to be 14.1 in., the absolute error is less than 1/10 of an inch. The relative error is less than (1/10) -^ 10 \/2 = 1/141, approximately, le. less than 0.71 per cent. 158. Rounded Numbers. Significant Figures. When the result of a measurement is expressed in the decimal notation, a generally adopted convention makes it possible to determine the degree of precision of the measurement from the number of significant figures contained in the number expressing the measure. This convention simply specifies that no more digits shall be written than are (probably) correct. Thus a measure- * The absolute error is therefore positive or negative according as the ob- served value is too small or too large. 236 IX, § 158] NUMERICAL COMPUTATION 237 ment of a length, expressed as 14.1 in. means that the measure is exact to the nearest tenth of an inch. If on the other hand the measurefment of this length were exact to the nearest hundredth of an inch, the measure would have been expressed by the number 14.10.* We should note, then, that the two numbers 14.1 and 14.10 do not mean precisely the same thing, when they express the result of a measurement. Again we may note that the absolute errors involved in the expression 4371.52 ft. and 42.81 ft. are each less than one hundredth of a foot ; whereas the relative error is in the first case less than 1/437152 and in the second only less than 1/4281. Sometimes we are furnished with numbers expressing meas- ures which are given with greater accuracy than we can use, or care to use. Thus suppose we want to express a measured length of 3.5 in. in terms of centimeters. We find in a table of equivalent lengths that 1 in. = 2.54001 cm. It would be obviously absurd to use this expression as it stands. We accordingly round it off to 2.54 or even to 2.5 and find that 3,5 in. = 8.9 cm. If, on the other hand, we wish to express 3.50000 in. in centimeters, we should have to use the value 2.54001. A number is rounded off by dropping one or more digits at the right, and, if the last digit dropped is 5% 6, 7, 8, or 9, in- creasing the preceding digit by l.f Thus the successive approximations to tt obtained by rounding off 3.14159 .- are 3.1416, 3.142, 3.14, 3.1, 3. * In other words x = 14.1 means that the exact value of x lies between 14.05 and 14.15; and x = 14.10 means that the exact value of x lies between 14.095 and 14.105. t In rounding off a 6, computers use the following rule : Always round off a 6 to an even digit. Thus 1.415 would be rounded to 1.42, whereas 1.445 would be rounded to 1.44. The reason for this rule is that, if used con- sistently, the errors made will in the long run compensate each other. 238 MATHEMATICAL ANALYSIS [IX, § 158 The significant figures of a number may now be defined as the digits 1, 2, 3, -, 9 together with such zeros as occur between them or as have been properly retained in rounding them off. Thus 34.96 and 3,496,000 are both numbers of four significant figures. On the other hand 3,496,000.0 has eight significant figures, since the in the first decimal place accord- ing to the convention adopted means that the number is exact to the nearest tenth. This zero is then essentially a digit properly retained in rounding off, and should be counted as one of the significant figures. Confusion can arise in only one case. For example, if the number 3999.7 were rounded by dropping the 7, we should write it as 4000 which, according to the rule just given, we would consider as having only one significant figure, whereas in reality we know from the way in- which the number was ob- tained that all four of the figures are significant. When such a case arises in practice we may simply remember the fact, or we can indicate that the zeros are significant by underscoring them, or by some other device. Computers adopt devices of their own to avoid errors in such cases. 159. Computation with Rounded Numbers. Addition. Since the (absolute) error of any approximate number can be at most one half the unit represented by the last digit at the right, the sum of n such numbers can be in error by at most n/2 times the unit represented by the last figure. These con- siderations lead to the following convention : in adding a column of approximate numbers first round off the given numbers so that not more than one column at the right is broken ; round off the sum so that the last figure to the right comes in the last unbroken column. This last figure is then uncertain. Nevertheless, it is usually retained temporarily. As a matter of fact, even the figure preceding this last one is IX, § 160] NUMERICAL COMPUTATION 239 not certain, since the errors may accumulate in adding several numbers. Eor example, in adding 21.64 3.8576 5.259743 10.31 we first round off : 21.64 3.858 5.260 10.31 41.068 = 41.07 The final sum is written 41.07, but even the last figure 7 is open to question. Show that the true result may be as low as 41.06 or as high as 41 .08. To retain all the figures in the second and third of the num- bers originally given would be absurd and would give in the result a misleading pretense of accuracy which does not exist in fact. In subtraction round off similarly. 160. Multiplication. Let a and b be approximate numbers and let their relative errors be a and ft respectively. The exact numbers are then (nearly) a -{- aa and b + bft. Their product is a6 + «6«+a&;8 + aJ«j8. The error committed in using ab as the product is then ab{a + ^ + «^) and the relative error is therefore nearly Now in practice a and /3 are small fractions, so that a ft is in- significant when compared with a + /8. (For example, if a and 240 MATHEMATICAL ANALYSIS [IX, § 160 /3 are both equal approximately to 0.001, ap is equal approxi- mately to 0.000001.) Hence, we conclude that the relative error of the product of two numbers is approximately equal to the sum of the relative errors of the factors. Hence, in finding the product of two approximate numbers, round off so that the two numbers have the same number of significant figures, and retain only this number of significant figures in the product* Even then the last figure retained may be unreliable. Example. Multiply 27.17 by^3, 14159. Round off the second factor to 3.142, and multiply: 27.17 X 3.142 5434 10868 2717 8151 85.36814 = 85.37 Even the figure 7 may be in error. Show that the true answer may be as low as 85.35. The labor involved in such a multiplication may be considerably re- duced by slightly modifying the method used, as follows : After having equalized the number of significant fig- ures annex a zero to the multiplicand. Multiply by the first figure on the left of the multiplier. Drop the last figure of the multiplicand and multiply by the second figure of the multiplier. Drop the next figure of the multiplicand and multiply by the third figure of the multiplier (but "carry" the amount from the figure dropped : thus .in the example having dropped the 7 and multiplying by 4, we say 4 x 7 = 28, carry 3, 4 x 1 = 4, +3 = 7, which is the first figure we write), and so on, arranging all the partial products so that the last figures from the left fail into the same vertical column ; then add in the usual way. * Since ^^ = 3.1428571, while tt = 3.1415927, the value y may be used for ir when the uncertainty of the other factors in a product in which it appears is greater than 1 part in 3000 (approximately) . 27.170 X 3.142 81510 2717 1087 64 85368 = : 85.37 IX, § 161] NUMERICAL COMPUTATION 241 161. Division. In case either the dividend (iV) or the divisor (D) is an approximate number, the following shortened method may be used : 1. Equalize the relative accuracy of N and D ; but if D is larger at the left, keep one extra figure on JV (as in the example below). 2. Divide as in long division, but drop successive figures in Z>, instead of adding successive zeros to N. Example. Find 295.679 -r- 7.53. (As 7 is greater than 2, we retain four figures in the dividend.) 7.53 1 295.7 [39.3 225 9 [3 X 753] 69 8 [divide by 75, gives 9] 67 8 [9x3 = 27, carry 3 ; 9 x 75 = 675, + 3 = 678] 2 [divide by 7, gives 3 (nearer than 2)] EXERCISES 1. Add the following numbers, each representing the result of a meas- urement: 25.62, 341.718, 2.62394, 28.7125. 2. Express 5.216 inches in centimeters. 3. Express 53.291 cm. in inches. 4. A rectangular table top is measured, and is found to be 2'4".5 X 3'6".4. Find its area. Find the error caused in this area if the measurements are each O'M too short. Find the relative error in the area. 5. Assuming that you can estimate the length and the breadth of a room which is about 15' by 18' to within 2', how nearly can you estimate its area ? 6. Assuming that you can measure each of the dimensions of the room of Ex. 5 with a yardstick to within 1" error, how nearly can you find the area of the floor ? If the height of the room is about 10', how nearly can you find the volume of the room by measurement ? 7. Assuming that you can measure the radius of a circle about 6" in diameter to within 0".l error, how nearly can you find its area ? How nearly could you find by measurement the volume of a cylinder about 5' high and about 5" in diameter ? 242 MATHEMATICAL ANALYSIS [IX, § 162 II. LOGARITHMIC SOLUTION OF TRIANGLES 162. Logarithmic Computation. We have already had occasion to observe that many computations in engineering, astronomy, etc., are carried out by means of logarithms. Jn the last chapter a few examples of the use of logarithms in computation were given in connection with a four-place table. Such a table suffices for data and results accurate to four sig- nificant figures. When greater accuracy is desired we use a five-, six-, or seven-place table. The methods used in connection with such a table differ slightly from those used ordinarily with a four-place table. Accordingly we take up briefly at this point some problems in- volving computation with a five-place table of logarithms. No subject is better adapted to illustrate the use of logarith- mic computation than the solution of triangles, which we shall consider in some detail. Five-place tables and logarithmic solutions ordinarily are used at the same time, since both tend toward greater speed and accuracy. 163. Five-place Tables of Logarithms and Trigonometric Functions. The use of a five-place table of logarithms differs from that of a four-place table in the general use of so-called " interpolation tables " or " tables of proportional parts," to facil- itate interpolation. Since the use of such tables of proportional parts is fully explained in every good set of tables, it is unnec- essary to give such an explanation here. It will be assumed that the student has made himself familiar with their use.* In the logarithmic solution of a triangle we nearly always need to find the logarithms of certain trigonometric functions. * For this chapter, such a five-place table should be purchased. See, for ex- ample, The Macmillan Tables, which contain all the tables mentioned here with an explanation of their use. IX, § 163] NUMERICAL COMPUTATION 243 For example, if the angles A and B and the side a are given, we find the side b from the law of sines given in § 125, , a sin B ' b = — — -' sin A To use logarithms we should then have to find log a, log (sin B) and log (sin A). With only a table of natural functions and a table of logarithms at our disposal, we should have to find first sin Ay and then log sin A. For example, if A = 36° 20', we would find sin 36° 20' = 0.59248, and from this would find log sin 36° 20' = log 0.59248 = 9.77268 - 10. This double use of tables has been made unnecessary by the direct tabulation of the logarithms of the trigonometric functions in terms of the angles. Such tables are called tables of logarithmic sines, logarithmic cosines, etc. Their use is explained in any good set of tables. The following exercises are for the purpose of familiarizing the student with the use of such tables. EXERCISES 1. Find the following logarithms : * (a) log cos 27° 40'.5. (d) log ctn 86° 53'. 6. (6) log tan 85° 20'. 2. (e) log cos 87° 6'. 2. (c) log sin 45° 40'.7. (/ ) log cos 36° 53'.3. (d) log sin A = 9,78332 - 10. (e) log ctn ^A = 0.70352. (/) log tan J ^ = 9.94365 - 10. 87325 4. Given a triangle ABC, in which ZA = 32°, ZB = 27°, a - 5.2, find b by use of logarithms. * Five-place logarltbms are properly used when angles are measured to the nearest tenth of a minute. For accuracy to the nearest second, six places should be used. 2. Find A , when (a) log sin A = 9.81632 - -10. (6) log cos A = 9.97970 - -10. (c) log tan ^ = 0.45704. 3. Find^, iftan. = 4^«-\2^ Q7Q. 89.710 244 MATHEMATICAL ANALYSIS [IX, § 164 164. The Logarithmic Solution of Triangles. The effective use of logarithms in numerical computation depends largely on a proper arrangement of the work. In order to secure this, the arrangement should be carefully planned beforehand by con- structing a blank form, which is afterwards filled in. Moreover a practical computation is not complete until its accuracy has been checked. The blank form should provide also for a good check. Most computers find it advantageous to arrange the work in two columns, the one at the left containing the given numbers and the computed results, the one on the right contain- ing the logarithms of the numbers each in the same horizontal line with its number. The work should be so arranged that every number or logarithm that appears is properly labeled ; for it often happens that the same number or logarithm is used several times in the same computation and it should be possible to locate it at a glance when it is wanted. The solution of triangles may be conveniently classified under four cases : Case I. Given two angles and one side. Case II. Given two sides and the angle opposite one of the sides. Case III. Given two sides and the included angle. Case IV. Given the three sides. In each case it is desirable (1) to draw a figure representing the triangle to be solved with sufficient accuracy to serve as a rough check on the results ; (2) to write out all the formulas needed for the solution and the check ; (3) to prepare a blank form for the logarithmic solution on the basis of these formulas ; (4) to fill in the blank form and thus to complete the solution. We give a sample of a blank form under Case I ; the student should prepare his own forms for the other cases. IX, § 165] NUMERICAL COMPUTATION 245 165. Case I. Given two Angles and one Side. Example. Given : a=430.17, ^=47° 13'.2, B=d2° 29'.6. (Fig. 134) To find: C, 6, c. Formulas : 6 = 180°- a sin J. a -(A+B), sin B, sin C. Check : Fia. 134 sin^ c-b ^ ts,ni(C-B) c + b tSinl{C-\-B)' The following is a convenient blank form for the logarithmic solu- tion. The sign ( + ) indicates that the numbers should be added; the sign ( — ) indicates that the number should be subtracted from the one just above it. ^ , , Logarithms A- Numbers ( + )^ = A + B = C = 179° 60' d — sin^ = sin • . a/sin A sin B = b = sin • • •) (+) ■) a /sin A sin C = sin c =.. . c-b = c+ b = (->) ( + ) Check C-B= C + B = . .... tan^C- J5) =tan . . . (- tan^(C-|-5)=tan . . . (- ■(1) (Logs (1) and (2) . should be equal .) ( — ) for check.) (2) 246 MATHEMATICAL ANALYSIS [IX, § 165 FilUng in this blank form, we obtain the solution as follows. Numbers Logarithms A = 47° 13'.2 B= 52° 29\6 A-\-B= 99°42'.8 179° 60'.0 C = 80° 17'.2 a = 430.17 (->) 2.63364 sin ^ = sin 47° 13' -2 (— >) (-) 9.86567-10 a/sin ^ 2.76797 sin B = sin 52° 29'. 6 (->) ( + ) 9.89943 - 10 b = 464.94 Ans. (<-) 2.66740 a/sin yl 2.76797 sin C = sin80° 17'.2 (->) ( + ) 9.99373 c = 577.70 Ans. (^^) 2.76170 c-6= 112.76 c + 6 = 1042.64 (->) 2.05215 (->►) (-) 3.01818 9.03402 - 10 C-B= 27°47'.6 C + 5 = 132° 46'.8 tan ^ ( a - jB) = tan 13° 53'.8 (->) 9.39342 - 1 tan 1(0+ -B)= tan 66° 23'. 4 (— ^) (_) 0.35942 9.03400 - 10 J Check EXERCISES Solve and check the following triangles ABC : 1. rt = 372.5, .4 = 25°30', J5 = 47°60'. 2. c = 327.85, vl = 110° 52'.9, B = 40° 31'.7. Ans. C = 28° 35'.4 a = 640.11, b =446.20. 3. a = 53.276, A = 108° 50'.0, C = 57° 13'.2. 4. 6 = 22.766, ^=141°59'.l, a=25°12'.4. 5. 6 = 1000.0, B = 30° 30'. 5, C = 50° 50'.8. 6. a = 257.7, A = 47° 25', B = 32° 26'. IX, § 166] NUMERICAL COMPUTATION 247 166. Case II. Given two Sides and an Angle opposite one of them. If A, a, b are given, B may be determined from the relation h sin A (1) sin B If log sin B = 0, the triangle is a right triangle. Why ? If log sin B >0, the triangle is impossible. Why ? If log sin B <0, there are two possible values Bi, B^ of B, which are supplementary. Hence there may be two solutions of the triangle. (See Ex. 1, page 249.) No confusion need arise from the various possibilities if the corresponding figure is constructed and kept in mind. It is desirable to go through the computation for log sin B before making out the rest of the blank form, unless the data obviously show what the conditions of the problem actually are. Example 1. Given : A = 46° 22'.2, a = 1.4063, 5=2.1048. (Fig. 135) To find: S, O, c. Formula : sin B b sin A a Fig. 135 Numbers & = 2.1048 (- sin ^ = sin 46° 22'. 2 (- bsinA a = 1.4063 (- sin B {< Logarithms 0.32321 ►) ( + ) 9.85962 ~ 10 0.18283 (-) 0.14808 .) 0.03475 Hence the triangle is impossible. Why ? 248 MATHEMATICAL ANALYSIS [IX. § 166 Example 2. Given : a = 73.221, b = 101.53, A ■. To find: J5, C, c. 40° 22'.3. (Fig. 136) Formula: sin 5 = 6sin^ Numbers b = 101.63 ( sin^=sin40°22'.3 ( b sin A a = 73.221 sin JB G Logarithms 2.00660 -) ( + ) 9.81140 -10 11.81800 - 10 1.86464 (-^) (-) 9.95336 - 10 The triangle is therefore possible and has two solutions (as the figure shows). We then proceed with the solution as follows : We find one value Bi of B from the value of log sin B. The other L JJ lO llUCJU j Other formulas : (7=180°-(^ + B). . asin C sin^ Check: ^-^: c + & tanUC-B) tanKO + 5) Numbers Logarithms sin B 9.95336 - 10 Bi= 63°65'.2 179*^ 60'.0 ^2 = 116° 4'.8 -4 + 5i = 104° 17'.6 179° 60'.0 Ci= 76'^42'.6 a (-» 1.86464 a'mA (-» (- -) 9.81140- a/sin A . 2.06324 sinCir=8in76°42.'6 '(->) ( + ) 9.98634-] Cl 10 10 109.64 (-^) 2.03968 IX, § 167] NUMERICAL COMPUTATION 249 Ci - 6 = 8.01 (. Ci + 6 = 211.07 (. Ci-^i= 11°47'.8 Ci + ^i = 139°37'.7 tan K Ci - Bi) = tan 6° 63'.6 tan K Ci + Bi) = tan 69=^ 48'.8 0.90363 .) (-) 2.82443 8.57920-10 9.01377 - 10 0.43455 8.57922 - 10 Check* One solution of the triangle gives, therefore, B=63° 55'.2, C = 76° 42'.5, c = 109.54. To obtain the second solution, we begin with B2 = 116° 4'. 8. We find C2 from C2 = 180° -(A + B^) ; i.e. C% = 2-3° 32'. 9. The rest of the com- putation is similar to that above and is left as an exercise. EXERCISES 1. Show that, given A, o, 6, if A is obtuse, or if A is acute and a > 6, there cannot be more than one solution. Solve the following triangles and check the solutions : 2. a = 32.479, 6 = 40.176, ^ = 37° 25M. 3. 6 = 4168.2, c = 3179.8, B = 51° 21'.4. 4. a = 2.4621, b = 4.1347, B = 101° 37'.3. 5. a = 421.6, c = 532.7, ^ = 49° 21'.8. 6. a = 461.5, c= 121.2, C = 22° 31'.6. 7. Find the areas of the triangles in Exs. 2-6. 167. Case III. Given two Sides and the Included Angle. Example. Given: a=214.17, 6=356.21, C=62°21'.4. (Fig. 137) To find : A, B, c. Formulas : 6 tan I {B- A) tan KB + A); b + a B+A= 180° - C = 117° 38'.6 ; _ gsin C _ 6 sin C sin A sin B * A small discrepancy in the last figure need npt cause concern. Why? 250 MATHEMATICAL ANALYSIS [IX, § 167 Numbers 6 - a = 142.04 ( 6 + a = 570.38 ( (b - a)/{h + a) tan ^{B + A) = tan 58° 49' .3 ( tan \{B- A) = tan 22° 22^2 ( .•.A= 36°27M 5= 81°11'.5 -) -) Ans. Ans. a = 214.17 (->) sin ^ = sin 36° 27'. 1 (->) a/sin ^ sin C = sin 62° 21 '.4 (->) c = 319.32 Ans. (^ Check by finding log (6/sin B). Logarithms 2.16241 (-) 2.75616 9.39625 - 10 ( + ) 0.21817 9.61442-10 2.33076 (-) 9.77389 -10 2.55687 ( + ) 9.94736 -10 2.50423 EXERCISES Solve and check each of the following triangles. 1. a = 74.801, h = 37.502, C = 63° 35'.5. 2. a = 423.84, 6 = 350.11, C = 43°14'.7. 3. 6 = 275, c = 315, ^ = 30° 30'. 4. « = 150.17, c = 251.09, ^ = 40°40'.2. 6. a = 0.25089, b = 0.30007, C = 42° 30' 20". 6. Find the areas of the triangles in Exs. 1-5. 168. Case IV. Given the three Sides. Example. Given : a = 261.62, b = 322.42, c = 291.48. To find : A, B, G. Formulas : s = l{a -\- b -\- c). y-J (« — q)(g-ft)(g-c) tan ^A= — ^ , tan I 5 = ** a s — b Check: A -\- B -\- C = 180°. tan I C = b=3gg.4g A Fig. 138 8 — C (§ 143) IX, § 168] NUMERICAL COMPUTATION 251 Numbers a = 261.62 6 = 322.42 c = 291.48 ( + ) (-) Logarithms 2s = 875.52 s = 437.76 s-a= 176.14 (->) s-b= 115.34 (-^) s-c = 146.28 (-^) 2.24586 2.06i98 2.16518 2s = 875.52 (Check.) 8 = 437.76 (->) 6.47302 2.64124 r s— a 3.83178 1.91589 2.24586 tan ^ ^ = tan 25° 4'.1 (<-) 9.67003 - 10 r s-b 1.91589 2.06198 tan 1 5 = tan 35° 32'.4 (<— ) r = s— c = 9.85391-10 1.91589 2.16518 tan ^ C = tan 29° 23'. 4+ (<-) A = 50° 8'. 2 ^ws. B= 71° 4'.8 ^ns. 0= 68° 46'. 9 Ans. 9.75071-10 179° 59'. 9 Check. EXERCISES Solve and check each of the following triangles : 1. a = 2.4169, b = 3.2417, c = 4.6293. 2. a = 21.637, & = 10.429, c = 14.221. 3. a = 528.62, b = 499.82, c = 321.77. 4. a = 2179.1, & = 3467.0, c = 5061.8. 5. a = 0.1214, 6=0.0961, c = 0.1573. 6. Find the areas of the triangles in Exs. 1-5. 7. Find the areas of the inscribed circles of the triangles in Ex. 1-6. 252 MATHEMATICAL ANALYSIS [EX, § 169 III. THE LOGARITHMIC SCALE — THE SLIDE RULE 169. The Logarithmic Scale. Let us lay off, on a straight line, segments issuing from the same origin and proportional to the logarithms of the numbers 1, 2, 3, 4, —. The base of the system of logarithms is immaterial. Let us label the end- points of these segments by the corresponding numbers. This gives a non-uniform scale, which is called a logarithmic scale. Such a scale is pictured in Fig. 139. [ T s 4 Fig. 139 rm A scale of this kind is easily constructed from the graph of the logarithmic function (Eig. 133). 170. The Slide Rule. The slide rule is an instrument often used by engineers and others who do much computing.* It consists of a rule (usually made of wood faced with celluloid) Fig. 140 along the center of which a slip of the same material slides in a groove. This slip is called the slide. The face of the slide is level with the face of the rule. * Engineers usually purchase rather expensive slide rules made of wood and celluloid. These are on sale in all stores which carry draftsmen's supplies. A very simple slide rule sufficiently accurate for class purposes is printed on hard pasteboard and is obtainable at reasonably small cost through any one of several manufacturers of instruments. Figure 140 is reproduced on a larger scale on the first fly-leaf at the back of the book. By cutting out this leaf and carefully cutting up the figure, a slide rule can be made by the student. This will not be very accurate, but it will suffice to illustrate the principles. XI, § 170] NUMERICAL COMPUTATION 253 Along the upper edge of the groove are engraved two loga- rithmic scales, usually labeled A and B, the scale A being on the rule, the scale B on the slide. (See Fig. 141.) The scales A and B are identical. The slide is simply a mechanical device for adding graphically the. segments on r 1 2 C I 5 6 7 8 9 1 ^ 1 I 1 1 I 1 , , , , 1 2 ^ ihiililijlilililiilililili [llllili m m m w'm T-n: ■] trhl 1 \ f v> 1 1 1 m lU ill 1 p M 1 1 :::nn]iiioi 1 \ C N t-n , ,,,,, ...^T-g; ■ 4 ■ , ) 6 7 8 9 I 2 3 y iTi"TiiiT-r 'IHW ]]Xii"":i "■""I""I] ^iiiiiiiiiiiiiiiiiiiiiiiiiii U.iuM.-ljLi ' 1 ..Mr.i:ii :ii:i4 1 2 J 4 Fig. 141 these scales. Since the segments represent the logarithms of the numbers found on the scale, the operation of adding the segments is equivalent to multiplying the corresponding num- bers. Thus, to find the product 2.5 x 3.2 move the slide to the right until the point marked 1 at the extreme left of the slide (scale B) is in contact with the point 2.5 on scale A (Fig. 141 shows the positions of scales A and B after this operation). The point 3.2 on scale B is then opposite the point 8.0 on scale A. The latter number is the required product : 2.5 X 3.2 = 8.0. A little reflection should make quite clear how the operation just performed is equivalent to adding the logarithms of 2.5 and 3.2 and then reading from the scale the number corresponding to the sum. We may note further that with slide set as in the example just worked it is set for multiplying any number by 2.5 ; i.e. every number of the scale A is the product of 2.5 by the number below it on scale B. The slide is therefore also set for division by 2.5. Every 254 MATHEMATICAL ANALYSIS [IX, § 170 number of scale B is the result of dividing the number above it by 2.5. Thus we read from the scale (set as before) that 7.2 -^ 2.5 = 2.9 approximately. Having now shown very briefly how the slide rule may be used for multiplication and division, let us examine it a little more closely. Scales A and B are labeled with the numbers 1,2, 3, 4, 5, 6, 7, 8, 9, 1,2, ...,9,1. It is natural to ask why the number following the 9 in the middle of these scales is not labeled 10 ? The answer is that the numbers on the slide rule are given without any reference to the position of the decimal point, just as the numbers in a table of logarithms are given without reference to the decimal point. The number 1 at the extreme left of the scale may represent either 1, or 10, or 100, or 1000, etc., or .1, or .01, or .001, etc. If the 1 at the extreme left of the scale represents 1, then the other numbers on the first half of the scale repre- sent 2, 3, ..., 9, the 1 in the middle represents 10, the 2 represents 20, and the successive numbers represent 30, 40, -., 100 (the last being represented by the 1 at the extreme right of the scale). If on the other hand the 1 at the left represents 100, the successive numbers represent 200, 300, ..., 900, 1000, 2000, ..., 10,000. If the 1 at the left represents .1, the successive numbers represent .2, .3, .••, .9, 1.0, 2.0, ••., 10.0 ; and so on. The reading of the subdivisions on the scales (A and B) should now offer little difficulty. Whenever an interval be- tween two successive numbers is divided by certain lines of the same length into 10 parts, each of these parts represents one tenth of the number represented by the interval in question. Thus, if we fix our attention on the division between 2 and 3, we note that a certain set of lines divides this interval into 10 parts ; if the 2 represent 2, these divisions represent respec- tively 2.1,2.2, ..., 2.9. On the other hand, if the 2 is thought IX, § 170] NUMERICAL COMPUTATION 255 of as representing 20, these divisions represent 21, 22, —, 29 : and so on. These divisions into ten are at some parts of the scale subdivided further into five or two parts. These parts then represent fifths or halves of the interval that represented a tenth. Thus we may readily locate on the scale the point representing 1.42 or the point representing 3.65. Turning our attention to scales C and D along the lower edge of the groove on the slide and the rule respectively, we note first that these two scales are also identical. Compar- ing them with scales A and B, we see that the unit chosen for C and D is just twice the unit of A and B. Hence the scales C and D can be used for multiplying and dividing just as scales A and B are used ; however on C and D our range is smaller. The range of numbers on A and B is from 1 to 100 ; on C and D only from 1 to 10. To make up for this limitation, scales C and D give greater accuracy. However, the principal reason for the existence of the second pair of scales is the fact that the two pairs of scales thus ob- tained furnish a table of squares and square roots. In view of the relation between the units with respect to which the two pairs of scales are constructed, every number of scale A is the square of the number vertically below it on scale D. Why ? In order that corresponding numbers on scales A and D may be accurately read off, every slide rule is provided with a runner, the vertical line on which connects corresponding numbers of the upper and lower scales. The runner also enables us to perform calculations consisting of several operations without reading off the intermediate results, thus saving time and securing greater accuracy in the final result. The actual use of the slide rule will be explained in the next article. The successful use of the slide rule depends largely on the ability to read the scales readily and accurately, accuracy 256 MATHEMATICAL ANALYSIS [IX, §170 often necessitating the estimating of numbers falling between the lines of division. The ability mentioned can be secured only by practice. A proficient operator, with a ten-inch slide rule, can always secure results accurate to three significant figures. This degree of accuracy is sufScient for many of the computations of applied science, manufacturing, etc., in which the slide rule is proving more and more useful. 171. The Use of the Slide Rule. All calculations in mul- tiplication, division, proportion, etc., are worked on scales Cand D unless the answer is so large that it does not lie on the scale. In that case scales A and B are used. Let us begin with pro- portion. On this topic, and on the corresponding property of the slide rule, all computations involving multiplication or division, or both, maybe made to depend in a very simple way. The property of the slide rule referred to is as follows : No matter where the slide be placed, all the numbers on the slide bear the same ratio to the corresjjonding numbers on the rule (due regard being had to the position of the decimal point). For example, if the slide be set so that 2 of O coincides with 4 of D, it will be observed that the same ratio 2 : 4 exists between every pair of corresponding numbers : 1 : 2, 3 : 6, 42 : 84, 125 : 250, etc. Explain why this is true. This leads at once to the rule for finding the fourth term of a proportion, when the first three are given. We give this rule in diagrammatic form, as follows : * To find the fourth term of a proportion : c D Set first term over second term. Under third term find fourth term. ♦ Iji this article we have followed to a considerable extent the treatment given in the Manual for the use of the Mannheim Slide Rule, published by the Keuffel and Esser Co., New York. IX, § 171] NUMERICAL COMPUTATION This gives the solution of the equation b X To find the product abj solve the proportion a X To find the quotient -, solve the proportion a__x The following examples will make clear the procedure. Example 1. Solve the proportion : 13/24 = 32/a;. 257 D Set 13 over 24 Under 32 find 69.1 Ans. Example 2. Solve the proportion : 13/24 = 75/x. Since the first two terms of the proportion are the same as in the pre- ceding example, we set the slide as before. We now find, however, that 75 on C is beyond the extremity of D. We accordingly set the runner on the left-hand 1 of 0, and then set the right-hand 1 of C on the runner. We find under 75 the number 138.5, the required value of x* (Justify the above use of the runner. ) The same example can be done on scales A and B with one setting, without using the runner. Example 3. Find the product: 23.2 x 5.3. c D Setl over 23.2 Under 5.3 find 123.0 Ans. Here we set the right-hand 1 on 23.2. Use whichever 1 serves. The decimal point, in this as in the other examples, is simply located by in- spection and a brief mental estimate of the answer. Here we see readily that the answer is something over 100; hence we locate the decimal point at the place to give us 123.0. * The .5 in this answer must be estimated. Usually, if more than three significant figures are obtained from the rule, the last is uncertain. 258 MATHEMATICAL ANALYSIS [IX, § 171 Example 4. Find the value o/364 -4- 115. c D Set 364 over 115 Find 3.17, Ans. over 1 Example 5. Find the circumference of a circle whose diameter is 42 ft. We multiply the diameter by tt = 3.14.* Hence, c D Set 1 over 3.14 Under 42 find 132.0 Ans. By ordinary multiplication we get 131.88 ; an example of the inaccur- acy of the fourth significant figure. Example 6. Find the continued product : 1.6 x 4.2 x 5.3 x 2,8. The abbreviation R. denotes the runner on the slide-rule. Set 1 over 1.6 R, to 4.2 1 to R. R. to 5.3 1 toR. Under 2.8 find 99.7 Ans. We add a fevnr more rules for computing various types of expressions involving scales A and B as vv^ell as C and D. (1) To find a^ xb: (2) To, A Find a2&. Ans. B over b. C Setl D over a ada'^-h b: A Find a2 h- b, Ans B Set 6 over 1. C D over a * The number t is usually raa>'ked on the scale. IX, § 171] NUMERICAL COMPUTATION 259 (3) To find geometric mean between two numbers a and b; i.e. lincl x, so that a/x = x/b. Let a < 6. A B Seta Below 6 C D over a findiX=G.M: (4) To reduce fractions to decimals Set numerator over denominator Find equivalent decimal above 1 These rules are not to be memorized. They will be used almost in- stinctively by one who has made the reason for each rule thoroughly clear to himself and who is in practice. EXERCISES 1. With a slide rule compute the value of : (a) 2.13 X 4.42. {h) 2,856,000 x 256,700,000. (&) 1.98x5.24., ___ 5,43^31.5 (c) 2.77 x 3.14 X 4.25. id) 8.27/2.63. (e) 5.48/3.26. (/) 10/3.14. {g) 0.000116 X 0.0392. (0 U) 21.4 7.64 X 4.14 21.2 67.4 X 25.5 X 19.7 4.64 X 18.4 2. With a slide rule compute the value of : (a) (2.85)2. (c) (1.86)3. (6) 3.72 X (2.23)2. (^) (6.24)2/26.3. 3. Find the circumference and the area of a circle whose radius is 4.16 in. 4. What is the length in feet of 27.3 meters, given that 26 meters = 82 feet ? Solve with one setting of the slide. 260 MATHEMATICAL ANALYSIS [IX. § 172 IV. LOGARITHMIC PAPER 172. Logarithmic Paper. Euled paper is printed, on which the rulings in both directions are spaced according to the logarithmic scale (§ 169), i.e. precisely as on a slide rule.* Such paper is called logarithmic paper. Samples of this ruling are shown in Figs. 142-143. 173. Plotting Powers on Logarithmic Paper. The graphs of equations of the type (1) y = kx"" can be plotted very readily on logarithmic paper. For, if we take the logarithms of both sides, we find (2) log 2/ = log A: -h n log x. Let us set Y=\ogy, K=\ogk, X=loga;; then (2) becomes (3) Y=K-\-nX. Now the equation (3) represents a straight line if X and Y be taken as the variables. This is precisely what happens if we plot the values of x and y from equation (1) on logarithmic paper ; for, when we plot a value for x on logarithmic paper, the distance from the left border is nothing else than logic, i.e. X; and similarly for Y. Moreover, the slope of the straight line represented by (3) is n, the exponent of x in (1) ; and the intercept on the Y axis is K= log k. Hence if values of x and y from (1) are plotted on logarithmic paper, the value of n in (1) appears as the slope of the straight line graph, and the value of k can be read off directly on the vertical axis. * On this account, it is possible to make a crude slide rule by using the edges of two sheets of logarithmic paper, sliding them along each other after the manner of a slide rule. IX, § 173] NUMERICAL COMPUTATION 261 Example 1. Draw the graph of the equation y = r^ on logarithmic paper. Take x — \, then y =\. Take x = 10, then y = 100. Plot these two points A (1, 1) and B (10, 100) (Fig. 142). Connect A and 5 by a straight line. This is the required graph. The graph may be drawn also by noticing that its slope is the exponent ifm 11,1111 1 1 1 1 1 1 11 -y!:::::::::: -: o II II 1 t 2 ___ Q 5 --- ::::::: :: t / ::::::: :: i f 3 -- 7 t .... 7 ::::::: ^ 1.5 2 L ::::::: ^E=- — ::::::: izzz zi 7 9 17 z ^ r^ ::::::7 :: / / J f __ t 'i _ La? Al 1.5 2 4 5 7 S91 1.5 4 5 6 7 891 Fig. 142 of X in the given equation, i.e. 2. Hence we may draw from A a line whose slope is 2. Show that this gives the same line, AB. We may use this graph to find squares or square roots. Thus, if a; = 4, we can note the point on the graph directly over 4, and read the_corre- sponding value of y, which is 16. Reversal of the process gives \/16 = 4. Likewise, if x = 4.5, we find y - 20.2+ ; and vl5 = 3.8, approximately. 262 MATHEMATICAL ANALYSIS [IX, § 173 Conversely, given a straight line on logarithmic paper, we know that its equation must be of the form (1). We can j&nd n by actually measuring the slope, and we can read off k on the vertical line through the point marked (1, 1), since if we place lOjYTTl — — 7 9 — t n 7 " / « ^ / J. / ■y " e= elongation in c i^r^pull in kg. y m. / i 2 ' e = .3 r J f 1.5 ^ ^ ' — --—< «J — / / _ TZ r ■y c / / / I ~~T k^-S,/__ 1 .2 = = .15 E i^----------A E 1 ■ 1 17 X .iLLLLL _J 1 ,15 .2 .3 .4 .5 .6 .7.8.91 1.5 Fig. 143 3 4 5 6 7 8 910 a; = 1 in equation (2), we have log?/ = logfc, whence 2/ = A;. Any other value of x may be used instead of £c = 1, but a; = 1 is most convenient because log 1 = 0. Example 2. A strong rubber band stretched under a pull of p kg. shows an elongation of e cm. The following values were found in an experiment : IX, § 173] NUMERICAL COMPUTATION 263 p 0.5 i:o 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 6.0 7.0 e 0.1 0.3 0.6 0.9 1.3 1.7 2.2 2.7 3.3 3.9 5.3 6.9 If these values are plotted on logarithmic paper, . it is evident that they lie reasonably near a straight line, such as that drawn in Fig. 143. By measurement in the figure, the slope of this line is found to be 1.6, approximately. Hence if we set P= logp, ^ = 'loge, we have ^=^+1.6P, where iTls a constant not yet determined ; whence loge = K-{- \.Q\ogp or . e = kp^-^^ where K = log k. If p = 1, e = A; ; from the figure, if j> = 1, e = 0.3 ; hence k = 0.3, and e = 0.3pi-6. EXERCISES 1. Plot on logarithmic paper the graph of each of the following equa- tions : (a) y — 7?. (c) y = afi. (e) y = S x^. (b) y = xi (d) y = x^-s. (/) y = 4.5 x^-^. 2. Draw the graph ot y = x-^. Note that the negative exponent — 2 gives simply what we ordinarily call a negative slope of — 2 for the straight line graph. 3. When air expands or is compressed (as in an air compressor) , with- out appreciable loss or gain of heat, the pressure p and the volume v are connected by the formula p = kv~^-^, approximately. Pressure is often measured in atmospheres, and volume in cubic feet. If we start with one cubic foot of air at one atmosphere of pressure, it is obvious that k = 1. Draw the graph for this case, and from it find p when V = 0.5 cu. ft. Find v when p = 5 atmospheres. Find v when p = 0.5 atmospheres. 264 MATHEMATICAL ANALYSIS [IX, § 173 4. The intercollegiate track records for foot races (1916) are as follows, where d means the distance run, and t means the record time : cl 100 yd. 220 yd. 440 yd. 880 yd. Imi. 2 mi. t 0:09| 0:21^ 0:48 l:54f 4 : lof 9:23f Plot the logarithms of these values on squared paper (or plot the given values themselves on logarithmic paper). Find a relation of the form t = k(P^. What should be the record time for a race of 1320 yd.? (See Kexnelly, Popular Science Monthly, Nov. 1908.) 5. In each of the following tables, the quantities are the results of actual experiments ; the two variables are supposed theoretically to be connected by an equation of the form y = fccC*. Draw a logarithmic graph and determine k and ?i, approximately : (a) (Steam pressure ; v = volume, p = pressure. ) V 2 4 6 8 10 p 68.7 31.3 19.8 14.3 11.3 (Saxelby.) (6) (Gas engine mixture ; notation as above.) V 3.54 4.13 4.73 5.35 5.94 6.55 7.14 7.73 8.05 p 141.3 115 95 81.4 71.2 63.6 54.6 50.7 45 (Gibson.) (c) (Head of water h, and time t of discharge of a given amount.) h 0.043 0.057 0.077 0.095 0.100 t 1260 540 275 170 138 (Gibson.) CHAPTER X THE IMPLICIT QUADRATIC FUNCTIONS Two-valued Functions I. THE FORMS Ax^-^ Ey+ C = AND By'^ -{- Dx -{- C = 174. The General Implicit Quadratic Function. We shall now return to the discussion of algebraic functions. We first discussed the explicit linear function y=mx-\-b, and the function y defined by the implicit relation Ax -f- By +0 = (Chapter III). Then we discussed the explicit quadratic function of the form y = ax"^ -\-hx -\- c (Chapter IV). We now propose to take up the discussion of the functions y defined by implicit quadratic relations, such as 4 !/2 — 5 a? = 0, ic^ — 4 ?/--f 2 a;— 41/— 1 = 0, etc. The most general form of such an equation is (1) Ax"^ + Fxy + By^ + Dx -}- Ey + C = 0. The graphs of equations of this form are important curves, with interesting geometric properties, which we shall discuss in a later chapter. Our present purpose is to determine the general nature of these graphs (their shape, etc.) and to develop methods whereby the graph of a given equation of the type considered may be readily drawn. We may note at the outset that the function defined by an implicit quadratic relation between x and y will usually be two-valued, i.e. to each value of x will correspond, in general, two distinct values of y. This is due to the fact that if any particular value be assigned to x in equation (1) above, the 265 266 MATHEMATICAL ANALYSIS [X, § 174 corresponding values of y are determined by a quadratic equa- tion, unless ^ = 0. We shall approach the discussion of equations of type (1) by considering in order certain simpler forms of this general type. First, we shall discuss equations of the two types Ax'+Ey +C = and By'^ ^ Dx -\- C =^ 0. 175. The Equations x'^ — y — and y'^—x—0. We can dis- pose of the equations x"^ — y = and 3/2 — a; = very quickly. The first equation is equivalent to the equation y = x^, already discussed in § 72. The second equation is equivalent to the equation (2) y^=x, or y = ± Va;. We can either plot the points {x, y) whose coordinates satisfy this relation and thus obtain the graph desired * ; or, we can note that the equation y"^ = x is obtained from the equation 352 = ?/ by simply interchanging x and y. Hence, the graph of y"^ = X is obtained from the graph of y = x^ by turning the plane of the graph oi y — x"^ over about the line through the origin bisecting the first and third quadrants. Eor, this opera- tion will interchange the x- and ?/-axes in the desired way. The two graphs are shown in Fig. 144. Certain properties of the graph of the equation y"^ = x are at once evident from the form of the equation : The graph is symmetric with respect to the a^axis ; for, if a point (Ji, k) satisfies the equation, the point {Ji, — k) also satisfies the equation. Why ? The graph lies at the right of the aj-axis ; for, any negative value of x would give rise to imaginary values, of y. Why? - * A table of square roots will facilitate the work. X, § 176] IMPLICIT QUADRATIC FUNCTIONS 267 I i i t ,i / 3 :_:_V-- — ^ :: = =— - 5 , S 1 ^'^^"' \ 7-A^ \ ^/1v_ __ __ _._ _. - - 1^ ix^ ::fe^ . ir 2' ■■ ^"^^L ^s 4^5; " ~ : J-"--.^ : r ._ ..,,, ^f^ "--^ . . __±i_J 1 1 1 1 M 1 1 M 1 i 1 11 1 1 1 M 1 m 1 M Fig. 144 The most important properties of the double-valued function ± Va; to be noted are the following : (1) For every positive value of x there are two values of the function, viz. -f V^ and — ^x. Therefore the function is two- valued. (2) As X increases numerically, the corresponding values of ■\Jx increase numerically, i.e. the numerical value of Va; is an increasing function of x. 176. The Form By'' -f Dx = 0. B^^. may always write the equation in the form (3) Since B^^^ we B ' Y ^ - n>o X y' i.e. in the form where n = — D/B. The graph is then similar to that of x^^ny, the only difference being that the roles of the X- and i/-axes are interchanged. If the coefficient n is positive, the graph is at the right of the 2/-axis ; if n is nega- tive, the graph is at the left of the .v-axis (Fig. 145). In both cases the graph is symmetric with respect to the ic-axis, and Fig. 145 268 MATHEMATICAL ANALYSIS [X, § 176 passes through the origin, at which point it has a vertical tan- gent. Why ? The curve defined by an equation of the type considered is called a parabola if D =^0. (See Chapter IV.) To sketch such a curve rapidly, knowing its general shape, we need only plot a few corresponding values of x and y. If i>=0, the equation becomes By'^=0. Its graph is then the a^axis. 177. The Slope of the Curve By^ + Dx = 0. To determine the slope of the tangent to the curve By^-{- I)x= 0, we may proceed by the method used for similar problems in Chapters IV and V. To this end we first calculate the change ratio Ay /Ax, which is the slope of the chord PQ (Fig. 146). The z^ \^Au TTo Fig. 146 slope of the tangent at P is then the limit which this ratio approaches when Ax approaches the value 0. Let P(£Ci, 2/i) be any point on the curve, and Q(a;i + Ax, t/i + Ay) be another such point. Then we have J5(2/i + AyY + i)(a!i + Ax) = 0, and Expanding the first of these equations, and subtracting the second from it, we get 2 By Ay + BAy'^ + DAx = 0, or {2By,-rBAy)^^=-D. X, § 177] IMPLICIT QUADRATIC FUNCTIONS 269 Hence, the desired change ratio is Aj/_ D Aa? 2Byi-\-BAy° When Ax approaches zero, Ay also approaches zero. Why? The desired slope of the curve Bi/ + Dx = at the point (xi, y-^ is, therefore, ^^ 2By, The expression for the slope exhibits certain properties of the curve : (1) The curve has a vertical tangent at the origin (2/1 = 0). (2) The slope of the curve above the it'-axis is positive, if B and D have opposite signs ; and negative, if B and D have the same sign. (3) The slope of the curve decreases indefinitely in absolute value as the point (a^i, 2/1) recedes indefinitely from the origin. EXERCISES 1. For each of the following equations, determine the slope at the point (xi, y\) and sketch the curve represented. For each point plotted deter- mine the slope of the tangent and draw the tangent. (a) 2/2-4x^:0; (6) ?/2 + 2a; = 0; (c)4x2-3y = 0; ((?) 4 2/2 + 9x^0; (e) y^ = Qx. 2. Derive the equation of the tangent to each of the curves in Ex. 1 at the point indicated : (a)(l,2); (6)(_2,-2); (c)(-3,12); (d) (-4, - 3) ; (e)(6,6)'. 3. Show that the equation of the tangent to the curve y2 = 2 px at the point (xi, yi) on the curve is y\y =p (x + xx). 4. Draw the curves y'^ = nx for several different values of n on the same sheet of paper. It is suggested that the values w = l,2, 5, — 1, — 2, be included. 270 MATHEMATICAL ANALYSIS pC, § 178 II. THE FORM Ax"^ + By^ + C= 178. The Case A=^B. The Equation x"" -^ y^ = a\ It so happens that, if the units on the x- and ?/-axes are equal, we can interpret the left-hand member of this equation geometrically. For, it is evident from the figure (Fig. 147) that, under the Fig. 147 hypothesis of equal units, x^ -\- y^ is the square of the distance of the point (x, y) from the origin. Hence the equation (5) a;2^2/' = «^ states that the point {x, y) is distant a units from the origin. It follows that the points {x, y) satisfying this equation are all on the circle described about as center with the radius a, and conversely the coordinates of every point on this circle will satisfy the equation. The graph of the equation x^ -\-y'^= a^ is then a circle, if the units on the two axes are equal. If the units on the axes are unequal, the ordinates of the above circle must be shortened or lengthened in a certain ratio, according as the unit on the ^/-axis is less than or greater than the unit on the a^axis. In either case the graph of the equa- tion will be a closed curve. Throughout the remainder of this chapter, however, we shall assume, in order to fix ideas, that the units on the axes are equal. If A = B {AB ^ 0), the equation (6) A^-^Bf-^-Q^^ X, § 179] IMPLICIT QUADRATIC FUNCTIONS 271 may be written in the form x^ -\- y^ = — — » The graph of this equation is a circle, if — C/A is positive. If — C/A is negative, the equation has no graph, i.e. no pair of real values of x and y can satisfy it. If C — 0, tHe only point satisfying the equation is the origin.* 179. The Case A >0, B >0. Consider first the special case x"^ + 4t y- = 9. If we solve this equation for ?/, we have (7) y = ±^^9-x\ Now, we know from § 178 that the graph of the function (8) 2/ = ±V9^=^ is a circle with center at the origin and radius equal to 3. Y ^-^ '~~-^ ^ \ >^ "^i ' -~ O-liji X y> VV T "^^ ^^ Z . , X .^ ■^^ ^--^ Fig. 148 The ordinates of the points of (7) are then equal to one half the corresponding ordinates of the points on the circle (8). The construction of the graph of (7) should then be clear from the figure (Fig. 148). The graph in question is a closed curve, having a greatest length of 6 units and a greatest width of 3 units. It is symmetric with respect to both axes. * The last locus may be considered as a circle with radius equal to 0; it is sometimes called a poini circle. 272 MATHEMATICAL ANALYSIS [X, § 179 (9) The general form Ax^-^ By^-j-C=0 can be treated similarly, if A and B are both positive, equation may be written in the form C A The (10) x^ + ?f = This shows that there is no graph if the right-hand member is negative. If the right-hand member is 0, the point (0, 0) is the only point satisfying the equation. There remains only the case where — C/A is positive. Equation (10) gives _ (11) ,=±^|.^_|_., Now, the equation (12) y represents a circle. Equation (11) tells us that the desired graph is obtained by shortening or lengthening the ordinates of this circle in the ratio ^A/B to 1. I >^fTTNL I I I I I I Example. If we solve the equation 9x^ 7^^1l^^tIIII -\-4y^ = S6 for?/, we obtain y = ± |V4 — x^; /[ I mI I I r\ I I I I this tells us that the graph of the given equation is obtained from that of the circle y = ± V4 — x'^ by lengthening the ordinates of the latter to three halves their original length. Figure 149 exhibits the result. mm The graph of an equation of the form Fig. 149 Ax^-\-By^-}-C=0 under the hypothesis that A and B are both positive and that C is negative, is then a closed curve symmetric with respect to both axes. The curve represented by an equation of the form (9) above is called an ellipse. An ellipse is symmetric with respect to X, § 180] IMPLICIT QUADRATIC FUNCTIONS 273 each of two perpendicular lines, called the axes of the ellipse. The intersection of the axes of an ellipse is called the center of the ellipse. Knowing the general shape of the curve, the quickest way to sketch it from the equation is to find the intercepts on the axes and draw a symmetric curve through the four points thus obtained. In the example 9 x^ -\- 4: y"^ == 36 already considered, we find the intercepts to be ic = ± 2 (found by placing 2/ = 0) and y = ±o (when x = 0). If we mark the four corresponding points, the curve can be sketched readil}. EXERCISES 1. Discuss the locus of each of the following equations and, if the equation has a locus, sketch it and show how it is related to a certain circle (if the locus is not itself a circle) : (a) x^ -\- y^ = 16. (d) ix^-\-y^-\-16 = 0. (g) ix^ -\- Sy^ = 12. (c) 4 0:2 4-2/2-16=0. (/) 2x^-{-2y2 = 5. ^ ^ 4 "^ 9 2. For what values of x in each of the equations in Ex. 1 doesy become imaginary ? For what values of y does x become imaginary ? 3. Show directly from the equations that each of the graphs in Ex. 1, if it exists, is symmetric with respect to both the x-axis and the y-axis. 4. According to the definition above, is a circle an ellipse ? 180. The Slope of the Curve represented by Ax^ + By^ -\- C = 0, Here again we calculate the change ratio Ay /Ax, which is the slope of the secant joining the points P(xi, y^) and Q{xi + Aa;, y^ + Ay) on the curve, and then find the limit which this ratio approaches when Q approaches P along the curve, i.e. when Ax and, consequently. Ay approach 0. The calculation is as follows : Since P and Q both lie on the curve , Ax' + By^+C^O, we have (13) Ax^^-{-By,^+C=0, T 274 MATHEMATICAL ANALYSIS [X, § 180 and (14) A{x, + Axy + B(y, + Ayf +C=0. Expanding the squares in the last equation and subtracting (13) from (14), we have 2 Ax^Ax + AAx"^ + 2 By^Ay + BAy^ = 0, or (2 By^ + BAy) Ay = — {2 Ax^ + AAx) Ax, whence we obtain the slope of the line PQ, Ay _ _ 2 Axi -f AAx Ax~ 2 By I + BAy {B=^0). When Ax and Ay both approach 0, we get for the slope of the curve at the point (xi, y^) (15) m = -^. « An interesting verification of this result may be noticed. It is well known that the tangent to a circle at a point P is perpendicular to the radius OP. Now consider a circle with center at the origin. The slope of the radius through (xi, yi) is then clearly yi/xi. The slope of the tan- gent should, therefore, be —Xi/yi. But this is exactly what the preceding formula for the slope gives, when the equation represents a circle, i.e. when A = B. EXERCISES 1. Show from the result of the last article that at the points where the curve Ax^ + By^ +(7=0 {ABC =^ 0) crosses the ?/-axis its tangents are horizontal ; and that at the points where it crosses the a;-axis its tangents are vertical. 2. Find the equation of the tangent to each of the following curves at the point indicated. Check the result by sketching the curve carefully and drawing the tangent from its equation, (a) 4 a:2 4- y2 = 25 at (2, 3). (6) x^ + iy^ = S&t (2, 1). (c) 3 x2 + 4 1/2 ::^ 16 at (2, -1). X, § 181] IMPLICIT QUADRATIC FUNCTIONS 275 181. The Case i4 > 0, 5 < 0. We may always write the equation (9) so that A is positive. The case where A and B have unlike signs leads to a new type of graph. The Graph of x^ — ]p- = 9. In seeking the graph of this equation, we observe first the following facts : (1) The graph crosses the a;-axis at the points (3, 0) and (—3, 0), and does not cross the y-axis. Why ? (2) The curve is symmetric with respect to both axes. For, if the point (/i, li) is on the curve, so also is the point (A, — H). Hence, the curve is symmetric with respect to the a>axis. Similarly, if the point (/i, fc is on the curve, so also is the point (— /i, Iz). Hence the curve is symmetric with respect to the 2/-axis. (3) Solving the equation for y gives us (16) 2/ = ± Va;2 _ 9, This incidentally again establishes the symmetry of the curve with respect to the x-axis. But it shows further that, if a'2<9, y is imaginary. Hence, no part of the curve lies in the strip of the plane between the lines x — Z and a; = — 3. In other words all values of x between 3 and — 3 are excluded. Solv- ing the equation for x gives a; = ± V2/' + 9. This shows that no values of y are excluded, since 2/^ + 9 is positive for every real value of y. (4) The slope of the curve at the point (a^i, y-^ is by § 180, m =— • 2/1 This shows that the curve crosses the a;-axis vertically, i.e. the lines a; = 3 and aj = — 3 are tangent to the curve at (3, 0) and (— 3, 0) respectively. With these results in mind we now calculate the coordinates of a few points on the curve and the slope of the curve at these 276 MATHEMATICAL ANALYSIS points. We thus get the following table : [X, § 181 X 3 4 5 6 y V7 4 3V3 m QO fvy f fV3 We plot these points and those symmetrically situated with respect to the two axes and get Fig. 150. We know from equation (16) that y increases numerically from as a; increases : :: :::::;: : : x::: -f- -- x ' ^ ~ G J2 A-:^ _ ^ TLt- : "s \ r, Ay "sZs ~ " ^ y " _s^s_ __; ^ _ _ <L I % \ z ^ I i : S ^ 5 /La \ ^ ^ ^ ' 5^1, it L S ^ Z J \ ^ 1 ^ i A si / t _ j: sc dt 1 ^ V i. aL 6 'C i zj's Jl t ^ ^5 y -/ - S^ ^L t ^ S 5 J /' S s - ^- z : :s^^. .-^t.^ s ^^ 2 2 - - ^s S: """ t -,Z_ \i s z:? _ _ s s ^/ - - ^^ z_ - . ^ i Fig. 150 numerically from 3. We have already seen that the curve consists of two branches. It remains only to consider what the character of the curve is for numerically large values of x. Equation (16) tells us that y increases numerically without limit, as x increases indefinitely in absolute value ; i.e. the curve recedes indefinitely from both axes. It recedes, however, in a very definite way. For, consider the slope m of the curve at any point (aji, 2/1). From § 180 we have, for A = l, J5 = — 1, X, § 181] IMPLICIT QUADRATIC FUNCTIONS 277 m=^= ^ Vi ± Va^i^ - 9 the upper sign being used if yi is positive ; the lower, if y^ is negative. To fix ideas, let (x^, yi) be a point in the first quad- rant and let it move out along the curve indefinitely. We de- sire to see what happens to the slope of the curve under this condition ; i.e. when Xi becomes indefinitely large. To this end we write m in a more convenient form, as follows : 2/ -2/] =-{^-^i)y which shows that as Xi increases indefinitely, m approaches more and more nearly the value -|- 1. This shows that the further the point {x^, y^) travels out along the curve in the first quadrant, the more nearly does the direction of its motion make an angle of 45° with the ic-axis. Consider now the equation of the tangent to the curve at the point {xi, ?/i) : 2/-2/] = 2/1 or, ^1^ - 2/i2/ = ^i^ - Vi^ or. This may be written xi 9 y =— ' X 2/1 2/1 As Xi and yi become indefinitely large, the slope Xi/yi, as we have seen, approaches the value + 1, while the term 9/2/i evi- dently approaches the value 0. Therefore, the tangent to the 278 MATHEMATICAL ANALYSIS [X, § 181 curve at the point {xi, 2/1) approaches the line y=:x. A line, which is the limiting position which the tangent to a curve approaches, as the point of contact recedes indefinitely along an infinite branch of the curve, is called an asymptote of (lie curve. If the point (x^, y^ recedes indefinitely along the curve in the third quadrant {x^ < 0, 2/1 < 0), the slope is positive and the tangent approaches the same limiting position as before, namely, y = x. Similar considerations (or the symmetry of the curve) show that the line y = -x] is also an asymptote. The two asymptotes are also shown in the figure as they are a great help in drawing the curve. The Graph of x"^ — y"^ If, in place of the 9 in the equation x"^ — y'^=^ just considered, we have any other positive number, say a^, the discussion is very similar and accordingly we can be brief. The curve of the equation x'^—y'^ = a^ crosses the pr- axis at the points (a, 0) and (—a, 0), and does not cross the 2/-axis. It is symmetric with respect to both axes. We have y = ± -y/x^ — a^ and m = Xi/y^. We find also 1 m= ± Fig. 151 4 from which we conclude that the curve approaches indefinitely near the straight lines y—x and y=—x. The curve is, then, as drawn in Fig. 151. X, § 181] IMPLICIT QUADRATIC FUNCTIONS 279 The General Case, when C is Negative. Any equation of the form where A is positive and B and C are both negative, may now be treated without much difficulty. Any such equation can be written in the form (17) x^ - ny = a\ From this we obtain n a\ Fig. 152 But this shows at once, by com- parison with the last equation considered, that the ordinates of points on the curve x^—n^if-^d?- are to the corresponding ordinates of the curve ^ — 'ip- =. o? as X/n is to 1. In Fig. 152 we have drawn both the curve a;2 — 2/2 == d?- and the curve x"^ — ^y"^ = a^, the ordinates of the latter being just one half of the corresponding ordinates of the former. The asymptotes of the latter are the lines y^^x and y = -^x. Since the asymptotes are a great help in sketching the curve, we should have a means of obtaining their equations quickly from the equation of the curve. From the result of § 180 (A=l, B=— n^) and considerations similar to those used in the discussion of x'^—y^=9, we find the equations m. the asymptotes to be y =- X and y = x, n n OT X— ny = and x-\-ny = 0. But these equations are found by placing equal to zero each of the factors of the left-hand member of the equation of the curve x^ — n^y^ = a^. 280 MATHEMATICAL ANALYSIS [X, § 181 An example will show how these various results may be applied in sketching a curve whose equation is of the form considered. To sketch the graph of 4iX^ — 9y'- = 36, we draw first tlie asymptotes 2x—3y=0 and 2 X + 3 y = (Fig. 153). We next place y = 0,in the given equation and find the a;-intercepts to be x = 3 and £c = — 3. We can now sketch the curve with considerable accuracy, since we know what its general charac- teristics are. ""■•Jnxrr t ;lL^' " : ■ -hi: :: ± :; : _;<?-- : S S T , 4^ , - 4- ^ ^s -( ^^ / - :_:--:±:j:___± i^.-zzt'.:.'- : :i::::±:±::i^?.-±{3. .± J /:^5-_^ \ ::::::±^i^::.::::-::!s-U:±:i:: ---^5"^-:-::::-::::::^^--^ /''^(^ _ - - ■^s*>. lX 4Ur<Tl T>T*sj I Fig. 153 The graph of any equation of the form ^2 _ „2^2 3= ^2 {^n ^ 0, a ^ 0) is a curve called a hyperbola. We have seen that it consists of two branches ; it is symmetric with respect to each of two lines, which are called the axes of the curve. One' of these cuts the curve in two points and is called the trans- verse axis ; the other axis does not meet the curve at all. The intersection of the axes of the curve is called the center of the curve. The branches of the curve extend indefinitely and approach two straight lines, the asymptotes of the curve, which pass through the center. We may now complete the discussion of the graph of any equation of the form Ax^ -\- By^ -\- C = 0, under the hypothesis that A is positive and B negative. We have already disposed of the case < 0, by considering the form x^ — 7iy — a,\ The case C > leads similarly to the form x^—n^y'^=—a^. By interchanging x and y this reduces to the form n^x"^ — y^ = a^ which on division by n^ reduces to the case O < already con- sidered. The graph of an equation Ax"^ -\- By"^ -\- C=0, when A is positive, B negative, and G positive, is therefore a hyper- bola with the center at the origin and with its transverse axis coinciding with the 2/-axis. X, § 182] IMPLICIT QUADRATIC FUNCTIONS 281 -^.= ^ =^^^- iii^?^-i:-S?"-:i - _ -l:^^Z ^^ ^^12] r -P^$ '^ T T l^h Fig. 154 It a ^^ ^-^ ^ » ^'^ ^v y^ :_ _i ^-.^^^ ' =^'^ ±s. " -,«: _ i Fig. 155 The following example will illustrate the method of sketching the curve : Sketch the graph of ic^ — 4 ?/2+ 4 = 0. The asymptotes are x — 2 y = and X + 2 1/ = (Fig. 154). Placing x = 0, we find the ^/-intercepts to be +1 and — 1. Having marked the corresponding points and drawn the asymptotes the graph is readily drawn. Finally, when 0=0, the equation may be written in the form x^—7i'^y^=0. This may be written (x — ny){x + 7}y) = 0. This equation will be satisfied by all points which satisfy either x—ny — or ic-f-7?y = 0, and by no others. The locus of the equation is then two straight lines passing through the origin. Figure 155 shows the locus of the equation 4 a;^— 9 2/^=0. 182. The Case il = or 5 = 0. If ^ = 0, 5 > 0, the equation Ax^-{-By^-\-C=0 becomes By^ -\- C — 0. If O is positive, there is no graph. If C is negative, the graph consists of two lines parallel to the £c-axis. If C is zero, the graph is the x'-axis. When B—OjA>0, the graph of the equation consists similarly of two straight lines parallel to the t^-axis, if C is negative ; of the ?/-axis, if is zero ; and there is no graph, if G is positive. EXERCISES 1. Sketch the graph of each of the following equations : (a) x2 - 9 ?/2 = 16. (d) 9 x'^ - 16 ^2 _^ 16 = 0. (g) 3 x2 - 2 y^ = 6. (&) x2~9i/2=- 16. (e) 9x^- 16?/2- 16 = 0. (/i) 3x2 - 12 = 0. (c) x2 - 9 2/2 = 0. (/) 9 x2 -'l6 1/ = 0. (0 3x2 + 1 = 0. 2. Give a detailed discussion of the graph of the equation x2— y^ =— 9 (analogous to the discussion of x2 — y2 _ 9 given in the text). 3. Give a detailed discussion of the graph of x^—n^y^=—a^. Prove, in particular, that the asymptotes of this hyperbola are given by x^—7i'^y^=0. 4. Prove that no tangent to the curve x^ - y^ = a^ has a slope that lies between + 1 and — 1. Prove, in general, that no tangent to the curve a;2 _ n22/2 = a2 (a =56 0) has a slope that lies between 1/n and — 1/n. 282 MATHEMATICAL ANALYSIS [X, § 183 III. THE FORM Ax^ -\- By'' + Dx -^ Ey + C = 183. Recapitulation and Extension of Previous Results. We have seen in the previous sections of this chapter that an equation of one of the forms By^ + Da; = 0, or ^a;2 -1-52/2 4- 0=0 represents either (a) a parabola, with vertex at the origin and axis coinciding with the oj-axis or the 2/-axis ; or, (h) an ellipse, with center at the origin and axes coinciding with the axes of coordinates ; or (c) a hyperbola, with center at the origin and transverse axis coinciding with the a;-axis or the y-axis ; or (d) two straight lines (which may coincide) ; or (e) a single point (the point (0, 0)) ; or (/) no locus. If we replace xhj x — h and yhjy — k, in any of the above forms, we know that the graph of the resulting equation is ob- tained from the graph of the original equation by moving the latter so that the origin moves to the point {h, k) (the axes re- maining parallel to their original positions). We may then conclude that an equation of any one of the forms (18) A{x - hy+ E{y - k) - 0, B{y - ky+ D{x -h) = 0, or A(x - hy -\-B(y-ky-\-C = represents either (a) a parabola with vertex at the point {h, k) and axis coin- ciding with the line x — h = or the line y — k = 0; or (b) an ellipse with center at the point (h, k) and axes coin- ciding with the lines x — h = and y — k = 0; or X, § 183] IMPLICIT QUADRATIC FUNCTIONS 283 (c) a hyperbola with center at the point (/i, k) and transverse axis coinciding with the line a;— /i=0, or the line ?/— A:=0; or (fZ) two straight lines (which may coincide), or (e) a single point (the point (h, k)) ; or (/) no locus. Now, any equation of the form (19) Ax' -\-Btf + Dx + Ey-\-C=0 can be put in one of the forms (18) by completing the squares. The following examples show how this may be done. axis'" ^ '^. Fig. 156 Example 1. Discuss and sketch the graph of y2_2y-f2x + 7 = 0. This equation may be written in the form y2_2y=-2x-7, or l/2_2y + l=-2x-7 + l, i.e. It is accordingly a parabola with vertex at ( - 3, 1 ) and axis y = 1. The graph is given in Fig. 156. Example 2. Discuss and sketch the graph of a;2 + y2 _ 4 a; _ 6 y + 9 = 0. This equation may be w^ritten in the form (x2-4x + 4) + (2/2-6?y + 9) =-9 + 4 + 9, or (a;-2)2 + (?/-3)2 = 4. Therefore the given equation represents a circle with center at (2, 3) and radius equal to 2. (See Fig. 157.) • Fig. 167 1, C _ 7 , i''----''^:- Hi" ? ^ ■? 1 ^ 284 MATHEMATICAL ANALYSIS [X, § 183 Example 3. Discuss and sketch the graph of 9x^ + IQy'^ — ISx + 64 2/ - 8 = 0. This equation may be written in the form 9(x2_2x+ )+16(?/2 + 4?/+ )=8, or 9(a;2-2a:+l)4-16(?/2+4?/ + 4) =8+9 + 64=81, i,e, 9(x-l)2+16(?/ + 2)2 = 81. - ■ ■ i, ...,.-- ffffiMffl™ ::: i::f :::q:: ::::: S 41 - T - ( - \ ± iL / - : "a ±i T "'-rr^ S -- It - - . 3 It : ± it Fig. 158 Hence this equation represents an ellipse whose center is at (1, — 2) and whose axes coincide with the lines x = l,y = — 2. The remainder of the discussion is left as an exercise. The graph is given in Fig. 158. Example 4, Discuss and sketch the graph of 9x2-36x-4?/2 + 24?/ = 36. This equation may be written in the form 9(x - 2)2 _ 4(y - 3)2 = 36, which is a hyperbola whose center is at (2, 3) (Fig. 159). It is left as an exercise to com- plete the discussion and prove that the equa- tions of the asymptotes are 3(x — 2) + Fig. 159 2(y - 3 ) = and 3(a; - 2) - 2{y - 3) = 0. ■ ■sry' ■ ■ " ^11-. ^ - ----- 5 " v" --A-' '- ■I /ti : ::: : -s:::: ni; : ::: if / li j/ N \ ^ j^. .... 1 :::. ::. :±.: _:±:::;__: EXERCISES Discuss and sketch the graph of each of the following equations 1. a;2 + 42/ + 4 = 0. 2. x2 + ?/2 + 4 X - 8 y + 1 = 0. 3. x2 - ?/2 + 2 X = 0. 4. x2 - 4 X + 2/2 + 2 ?/ + 1 = 0. 5. x2 + 4 X + 2 2/2 + 4 y + 1 = a. 6. 9 x2 + 4 2/2 - 36 X - 8 2/ + 4 = 0. 7. 9x2-42/2- 36x + 82/ = 4. 8. 2/2 + 22/- 12x- 11 =0. 9. x2 + 15 2/2 + 4 X + 60 ?/ + 15 = 0. 10. x2 - 3 2/2 - 2 X - 6 2/ + 7 = 0. X, § 184] IMPLICIT QUADRATIC FUNCTIONS 285 184. The Slope of the Curve Ax^ j^ By'^ -\-Dx -{-Ey + C = 0. Let P(aJi, 2/i) ^^^d Q(Xi -|- ^x, y^ + A?/) be any two points on the curve. Then Ax,^ + By,^ + Dx, + ^2/i + C = 0, ^(a^i -h Ax-)2 + 5(2/1 + Ay)2+ i>(a;, + Aa;) + E{y, + A^/) + C = 0. Expanding the second of these equations and subtracting the first from it, we have (2 Ax, + A^x + B)^x -I- (2 %i -f- SAt/ 4-^) a?/ = 0. Therefore the change ratio, or the slope, of the secant PQ, is A^ _ _ 2 Ax^ -\- A^.x + D A.i-~ 2 By, -\- B^y -\- e' If we let Aa; approach zero. A?/ will approach zero also. Why ? Therefore the slope of the curve at any point (Xi, y,) is 2Ax,-\-D m = —^ — • 2By, + E Example. Find the equations of the tangent and the normal to the curve x^ + 4y'^ — ix-\-2y — S = 3i,t the point (1, 1). Solution : The slope of the tangent at any point (xi, yi) is 8^1+2 At the point (1, 1) this slope is |. Therefore the equation of the tangent isy— l=i^(x— 1) and the equation of the normal is y — 1 = — 5(x — 1). EXERCISES 1. Find the slope of the tangent to each of the following curves at the point specified. (a) x2 + 2?/- 3 = at (1, l)j (b) x^-\-y^-4 = at (1, V3)^ (c) a;2-2 2/2 + 5 = at (1, V3); (d) 4x^-\-y^-2x-Sy-lO = at (2, 1). 2. Find the equation of the tangent to each of the curves of Ex. 1, at the point specified. 286 MATHEMATICAL ANALYSIS [X, § 185 IV. THE FORM Fxy + Dx + Ey + C = 185. The Graph of xy = a. The graph of the curve xy = a is symmetric with respect to the origin ; for, if the coordinates Qi, k) satisfy the equation, the coordinates {— h, —7c) also satisfy it. Since y = a/x, it is evident that x may assume all Fig. 160 Fig. 161 values except 0. (See § 36.) As x increases numerically without limit, the curve approaches the line y = 0, i.e. y =0 is an asymptote. Similarly as y increases without limit, the curve approaches the line a? = as an asymptote. It will be proved later that the curve is a hyperbola, provided a is not equal to zero. If a is positive, the graph is as in Fig. 160. If a is negative, the graph is as in Fig. 161. If a is zero, the graph consists of the two axes x = and y = 0. 186. The Graph of Fxy -\- Dx -\- Ey + C = 0. If in the equa- tion xy = a we replace x hy x — h and yhjy — k, we know that the graph of the resulting equation is obtained from the graph of the original equation by moving the latter so that the origin moves to the point (^, fc), the axes remaining parallel to X, § 187] IMPLICIT QUADRATIC FUNCTIONS 287 their original positions. It follows that the equation {x — h){y — k) = a{a^O) represents a hyperbola whose asymptotes are x = h, y = k. If a = 0, the equation represents the two lines x =^ h, y = k. Example. Discuss and sketch the graph of xy + 4x -{-2y = 1. First we write (ix±?){y±?) = l. Then from inspection we see that the given equation may be written in the form (x + 2)(2/ + 4)=9. That is, the graph is a hyperbola whose asymptotes are x = — 2, y = — 4. (See Fig. 162.) Fig. 1G2 187. The Slope of the Curve Fxy + Dx -^ Ey + C = 0. It is left as an exercise to show that the slope of the curve Fxy-^Dx-{-Ey-\-C=0 at any point (x^, y^ is Fy,A-D m = — -p— - • Fxi -+- E EXERCISES 1. Discuss and draw the graph of each of the following curves : (a) xy = l; (b) xy=-l; (c) xy = 2; (d) xy =- 2; 2. Discuss and draw the graph of each of the following curves. (a) xy + 2x = S; (b) xy + 2 x -\- iy = S; (c) xy - 4x + Sy =2, 3. Draw the family of curves xy = a, taking several positive and several negative values of a. How does xy = 0, compare with these ? 4. Show that any equation of the form ^ cx + d can be reduced to the form given in § 186. 288 MATHEMATICAL ANALYSIS [X, § 188 V. THE GENERAL FORM Ax^ + Fxy + By^ + Dx +Ey -\- C = 188. The Graph. Methods of drawing the graph of an equation in the above form will be illustrated by means of the following examples. Example 1. Discuss and sketch the graph of x^ + 2xy + y'-^ -2x-2=0. Solving for y, we have y = — x ±y/'2x -\- 2. All values of x less than — 1 must be excluded, for these values make 2 x + 2 negative. Similarly, since x=—(y—l)± V — 2?/ + 3, it follows that all values of y greater than I must be excluded ; for these values make — 2 2/ + 3 negative. The a;-intercepts are the roots of the equation a:- — 2 aj — 2 = 0, i.e. 1 ± y/S. The ?/-iritercepts are the roots of the equation y^—2=0, i.e. ± V2. From y —— X ± V2 X -{ 2 it is seen that x may start with the value — 1 and increase without limit. Similarly from x = — (y—1) ±V— 2y + S we see that y may start with the value f and decrease without limit. Using the above data and plotting the points Fig. 163 X -1 1 2 1±V3 y 1 ±V2 1,-3 -2±V0 we obtain the graph in Fig. 163. This, problem may be approached from an entirely different standpoint. Suppose we let y' = ± y/2 x-i-2 and y'f = —x. Plotting these curves* (Fig. 164), adding the ordinates of y' =±y/2x -\-2 to the ordinates ofy" = — x, gives us the desired graph. This may be done graphically. We have here a shear of y' HZ -t- v/2 x + 2 with respect to the line y" = Y ,CS^ it -4^ s \l \ ^D'^^ c ^^ lS- ^^ -J -^ ^^ ^^ Sv ^^ ^ \"^>sS v^5^ ^ ^ \ ^ 3_ _r Fig. 164 X. (See § 90.) * Observe that the equation ?/' =± v'2 a: + 2 is equivalent to y"^ = 2{x + 1), X, § 188] IMPLICIT QUADRATIC FUNCTIONS 289 Example 2. Discuss and sketch the graph of 2/2 - 2 xy + 2 x2 — 5 a; + 4 = 0. Solving for ?/, we have y = x± V— x^ -\- dx — 4:. Hence, we merely have to shear the circle 1/ -±V(x- l)(4-x), X- + y2 _ 5 a; + 4 = 0, with respect to the line y" = a: in order to obtain the desired result. (See Fig. 165.) The complete discussion is left as an exercise. Example 3. Discuss and sketch the graph of 7 a-2 + 36 xy -S6y^-25 = Solving for y, we have y = lx± iVl6a;2-25, Y. ^'^N -,' -X ^'^ Z -.2 y Z i -.^-Z I t-7^ J^ ^^v y^ Z^^ z2^^ \ -/ - _ __ _ /O X ^ \ J ^^^--^ Fig. 165 T7- 1 I J / ^ -^ - ^ " U A y> . ^> - t ,^^%r^ ^>s ^%^^-r^'i^ ^ N / X '^^^ i-C ^ SZ ^ !&^ ::::::_::n^^::::^^:^: :_:_:::: -::^:::^--:^^-:S-J4^^^p- ~ ) .^^"^ ^^'■' , *--U. _ 2ES'' 1 N,^. <^ N 5^^». yT^^i ^^l ' ^^^Jl^\r •' ^ss ^^ ^ - t ^5^ ^ J Si ?' ' it ^z : ^2 - 1 Fig. !()(> which shows that the desired graph may be obtained by shearing I.e. y=±iVl6x2-25, 16a;2_36i/2_25 = 0, with respect to the line y = \x (See Fig. 166.) The complete discussion is left as an exercise. 290 MATHEMATICAL ANALYSIS [X, § 188 EXERCISES Discuss and sketch the graph of each of the following equations : 1. Ax"^ + y"^ — i xy - X -\- S z= 0. 4. iy^ - ixy + x:^ = 1 - x. 2. 7f-2xy + 3x = 2. 5. Qy'^ -12xy -\-'Sx^ + ox = 6. 3. 2/2-8 xy +iex'^=l- x2. 6. y^ - 6xy -\- Sx^ - \0x - 25 = 0. 189. The Slope of the Curve Ax^ + By"^ + Fxy -\- Dx + Ey + C = 0. It is left as an exercise to prove that the slope m at any point {xi, y^) is . 2By,-^Fx,+E EXERCISES Find the equations of the tangent and the normal to each of the follow- ing curves at the points indicated. 1. 48 x2 - 11 a;y - 17 2/2 - 129 a: + 24 ?/ + 81 = ; (2, 1), (3, - 3). 2. x?/ 4- 2 X - a;2 4- ?/2 + 6 2/ = ; (0, 0), (0, - 6). 3. 81 y2 + 72 xy + 16 x2 - 96 x = 378 y - lU ; (3, 2). 190. A General Theorem. The results of the examples and exercises of § 188 suggest that the graphs of equations of the second degree involving an xy-term are similar to the graphs of equations of the second degree in which the xy-term is lacking. We may now prove that this is a fact. The theorem is as follows : Any equation of the form Ax^ + Fxy + By- + Dx -{■ Ey + C = repre- sents either an ellipse, or a hyperbola, or a parabola, or two straight lines {which may coincide), or a single point, or no locus. We shall prove this theorem by showing that if the locus of the equation (20) Jx2 + Fxy -{- By"^ + Dx + Ey -\- C=0 be rotated about the origin through sl properly chosen angle d, its equation will be of the form (21) ^'x2 + Bhf + D'x + ^'y + C = 0. The theorem then follows from § 183. X, § 190] IMPLICIT QUADRATIC FUNCTIONS 291 We saw in § 137 that, if any point P(x, y) be rotated about the origin through an angle ^ to a new position P'(a;', y'), the coordinates of P and P' are connected by the relations : x' =x cos d — y sin 6, xsin d -\- y cos d. (22) Solving these equations for x and y in terms of x', y\ we obtain X = x' cos d -\-y' sin d, ^ -^ y =— x' sin 6 -\- y' cos 6. If P(x, y) satisfies equation (20), P'(x', y') will satisfy the equation ob- tained by substituting the values of x, y from (23) in equation (20) . The result of this substitution is as follows : A (x' cos d -\- y' sin d)^+ F(x' cos^ + y'sin e){—x' sin d -{- y' cos 6) + B(— x' sin e + y' cos ey + D{x' cos d + y' sin d) + E(—x' sin e + J/' cos ^) + O = 0. When expanded and rearranged according to the terms in x', y\ we obtain (24) A'x'-^ + F'x'y' + B'yi'^ + D'x' + i^'?/' + C = 0, where A' = A cos^ + P sin2 ^ - P sin ^ cos 6. F' = 2(A- B) sin ^ cos ^ + P(cos2 6 - sin^ ^). B' = A sin2 ^ + P cos2 ^ + P sin 6 cos ^. D' = Bcos e— E sin ^. E' = D sin ^ + P cos ^. C" = C. Equation (24) will be of the desired form (21), if the angle 6 is so chosen that F' = 0. Now, F may be written (25) P' = (^-P)sin2^ + Pcos2^: F will, therefore, be equal to zero, if tan 2 ^ ^ B-A A value of 6 satisfying the condition (26) can then always be found.* This completes the proof of the theorem. The following exercises will illustrate the above proof. The method may also be used to draw the graphs of equations involving the xy-term. * If ^ = ^, we take 26 = 90°, i.e. d = 45°. 292 MATHEMATICAL ANALYSIS [X, § 190 EXERCISES Determine the angle ^through which the loci of the following equations must be rotated in order that their new equations shall contain no xy-term. Determine the new equation and use it to draw the locus of the original. 1. Sx^ + 4xy -\-5y'^-36 = 0. Solution : After substituting x =x' cos -\- y' sin ^, y z=— x' sin d 4- y' cos 6, the equation becomes (1) (8 cos2 ^4-5 sin2 6- 4: sin 6 cos e)x'^ + [6 sin d cos e-i- 4(cos2 d — sin^ e)'\x'y' + (8 sin2 6+ b cos^ ^ + 4 sin ^ cos 0)i/'2 _ 35 _ q. 2 tan e Therefore, tan 2 ^ = — 3 1 - tan2 d Solving this equation for tan 6, we have 4 tan2 ^ _ 6 tan — 4 = 0, or tan ^ = 2 or — ^. We choose tan 6 rant) ; therefore 2 V5' 2 (^ in first quad- . 5 ' . "T^ ^ ._ ^ - ^s ^ / \" -S ^"^ t ^' S.^J' jj ^ ^^ ji:_.i \^ \ 1 / ^'\ \ J^ ^^ A ^ mrcltan 2 "■ \ y' \ . Sin cos d . 1 Substitutin, obtain 2. x2 - ?/2 _|_ 2 xy - 12 = 0. 5 these values in (1) we 4 x'2 + 9 ?/'2 = 36. The desired graph is obtained from the grajjh of this equation by rotating it through the angle — Q about the origin. The construction of the adjacent figure explains itself. 5. 3 x2 - 2 a;?/ + 1/2 ^ 6 = 0. 12 = 0. 6. 8x2 12 xy + 3 y2 _ 36 = 0. 3 ?/2 + 42 = 0. 3. x2-y2_|.2x?/ + 2x 4. xy = 4. 7. 2 x2 - 12 xy 8. 6 x2 + 4 xy - ?/2 + 48 X - 12 y - 10 = 0. 9. 9 2/2 + a;2 + 2 xy = 0. 10. Prove that the locus ot xy = c may be rotated about the origin so as to coincide with the locus of x^—y'^ — a^, provided a^ =±2c. 11. With the notation of § 190, prove that A' -\- B' = A -\- B and that {A' - B')^ + F'-^ =(A- BY + F\ PART III. APPLICATIONS TO GEOMETRY CHAPTER XI THE STRAIGHT LINE 191. Introduction. We have hitherto used coordinates pri- marily for the purpose of representing functions graphically and investigating the properties of those functions. We have seen that every continuous function defines a curve or a straight line, the graph of the function. Thus far, we have laid emphasis only on the discovery of the characteristics of the functions from the known properties of the curves that repre- sent them. Conversely, we have seen that every curve or straight line, in the plane of a system of rectangular coordinates, defines a function ; i.e. the points of any such curve associate with every value of X one or more values of y. If this function can be determined when the curve is given, the properties of the curve may be studied from the properties of the function. This function is usually expressed by means of an equation in X and y, called the equation of the curve. We propose now to study the properties of various curves by means of their equa- tions. (See § 62.) Up to this time, we have used different scales on the two axes whenever it was convenient to do so. Throughout this and the next four chapters we shall assume, unless the contrary is specifically stated^ that the units on the x- and y-axes are equal. 294 MATHEMATICAL ANALYSIS [XI, § 192 192. The Distance between two Points. Given the two points Pi {xi, yi) and P2 {x2, y^)^ let us find the length of the segment PiP^- If a line be drawn through Pj parallel to the ic-axis and another through P2 parallel to the 2/-axis to form the right triangle P1QP2 (Fig. 167), we have at once (1) P,P,=^J\Q' + QPl T P. ^ N, Pr^ Q Mr M,'X Fig. Pi M, N, JU The segment PiQ is equal to the projection M^M^ of PxPi on the a^-axis and (^P^ is equal to the projection NiN^ of P1P2 on the 2/-axis. By the result of § 37, we have PiQ= M^M^ = x.^— ccj, QA=i^==2/2-2/i. Substituting these values in (1), we have the desired formula : (2) PxP,=V(x2-Xiy-j-{y,-yi)\ 193. The Simple Ratio. Given two distinct points Pj, P2 and any point P (distinct from P2) on the line P1P2, the ratio P1P/PP2 is called the simple ratio of P with respect to Pi, P2. The line-segments in this definition are directed segments. Accordingly the simple ratio of P with respect to Pi, P2 is positive if P is between Pi and P2, and negative if P is on either prolongation of the segment P1P2. XI, § 194] THE STRAIGHT LINE 295 194. Point of Division. The coordinates (x, y) of the point P on the line joining Pi (x^, y^ to P2{x2y 2/2) such that the simple ratio are given by the formulas (3) ;C=£l±J^, ^^J/l + Xyg. 1 + X l + X Proof. Draw lines through Pi, Pg, P parallel to the axes, meeting the jc-axis in Mi, M^, M, and the y-axis in Ni, N2, N, respectively (Fig. 168). Then, since P1P/PP2 = X, we have MiM ^ ^^=X MM^ ' NN, The first of these relations gives (by § 37) X2 — X Solving this equation for x gives Xi + \X2 r N Ml / P M ■^2 / X Pi N, Q Fig. 168 x = 1+A Similarly from the second relation above we obtain y = _ 2/1 + ^2/2 1+X The mid-point of P1P2 is obtained from the value A = 1. Why ? Accordingly the coordinates of the midpoint of P1P2 are f xi + X2 Vi + y2 \ l~2-' 2 ) 296 MATHEMATICAL ANALYSIS [XI, § 194 EXERCISES 1. Find the distance between the following pairs of points : (1, 2) and (5,3); (- 1, 6) and (2, -3); (-2, -1) and (-1,4); (-3,4). and (1,4). 2. Find the lengths of the sides of the triangle whose vertices are (—1, 1), (4, — 4), and (1, 3). Prove that it is a right triangle. [Hint : A right triangle is the only kind of triangle in which the square of one side is equal to the sum of the squares of the other two sides.] 3. An isosceles triangle has its vertex at (4, 4) and the vertex of one of its base angles at (0, — 1). The vertex of the other base angle is on the a;-axis. Find the coordinates of the latter vertex. [Hint : Let the unknown point be (ic, 0) and equate the equal sides. How many solutions are there ?] 4. Find the coordinates of the point whose simple ratio with respect to (2, 1) and (—4, 7) is 2. Find the coordinates of another point whose simple ratio with respect to the same two given points is — 2. Draw a figure illustrating this problem. 5. Check the result of Ex. 4 by calculating the lengths of the seg- ments involved. 6. Find the coordinates of the point which divides the segment from (2, — 1) to (— 4, 3) internally in the ratio 1 : 4. 7. Find the coordinates of the mid-points of the sides of the triangle in Ex. 2. 8. A quadrilateral has its vertices at the points (—2, 1), (3, 1), (5, 3), and (0, 3). Show that its diagonals bisect each other. What kind of a quadrilateral is it ? 9. Find the coordinates of the points of trisection of the segment from (3, -6) to (0, 3). 10. A triangle has its vertices at the the points (0, 4), (2, — 6), ( - 2, — 2) . Find the coordinates of the points two thirds of the way from each vertex to the middle point of the opposite side, and thus show that the three medians of the triangle all pass through the same point. 11. The vertices of a triangle are (xi, yi), {xi,y-2), (scs, yi). Find the coordinates of the point of intersection of the medians. 12. Show that the triangle ^1(4, 1), i?(l, 4), C(5, 6) is isosceles. XI, § 194] THE STRAIGHT LINE 297 13. One end of a line whose length is 13 units is at the point (3, 8). The ordinate of the other end is 8. What is its abcsissa ? 14. The middle point of a line is (2, 3) and one end of the line is at the point (4, 7). What are the coordinates of the other end ? 15. The points (2, 1), (3, 4), (— 1, 7) are the mid-points of the sides of a triangle. Find the coordinates of the vertices. 16. Find the area of the isosceles triangle whose vertices are (4, 1), (1, 4), (5, 5) by finding the length of the base and the altitude. 17. What equation must be satisfied if the points (a:, y), (2, 1), (1, 4) form an isosceles triangle the equal sides of which meet in (ac, y)? 18. Prove that the points (- 2, - 1), (1, 0), (4, 3) and (1, 2) are the vertices of a parallelogram. 19. The line from (xi, yi) to (^2, ?/2) is divided into 5 equal parts. Find the coordinates of the points of division. 20. A point is equidistant from the points (2, 1) and (—2, 1) and 7 units distant from the origin. Find its coordinates. QUESTIONS FOR DISCUSSION 1. Does the distance between two points depend on the order in which the points are taken ? Does the formula for the distance give the same result no matter in which order the points are taken ? Why ? 2. Does the simple ratio of a point with respect to Pi, P2 depend on the order in which the points Pi, P2 are taken ? What is the relation between the simple ratio of P with respect to Pi, P2 and the simple ratio of P with respect to P2, Pi ? [ Hint. The answer to this question follows most easily from the defini- tion of simple ratio. Prove the relation in question by means of the formulas in § 194. ] 3. Can the simple ratio of a point P with respect to Pi, P2 be — 1 ? Why ? As the simple ratio approaches —1 what is the motion of P? 4. What can be said of the position of the point P, if its simple ratio with respect to Pi, P2 is positive ? if its simple ratio lies between and — 1 ? if its simple ratio is less than — 1 ? 6. If the simple ratio of P with respect to Pi, P2 is X, what is the simple ratio of Pi with respect to P and P2 ? of P2 with respect to Pi and P ? 298 MATHEMATICAL ANALYSIS [XI, § 195 195. The Area of a Triangle. One Vertex at the Origin. Let us try to find the area of a triangle whose vertices are 0(0, 0), Pi(xi, yi), and P2(^'2j 2/2)- I^^t the angles XOPi and XOP2 be denoted by Oi and O2, respectively, and let the angle F1OP2 of the triangle have the absolute measure 6 (Fig. 169). Fig. 169 The area of the triangle is then equal to ^ OPi • OP2 sin 0. Now, the directed angle P1OP2 differs from 62 — ^1, if at all, only by multiples of 360° (§ 101). Therefore sin ^ = ± sin (PiOP2)= ± sin {$2 - ^1). The area of the triangle OP1P2 is, then, A=:±^OPi' OP2 sin(^2 - ^1) = ± I OPi ' OPsCsin $2 cos $1 — cos $2 sin ^1) (§138) = ±iOP,-OP, 2/2 _^i ^ Jh\ OP2 OPi OP2 OPj The area of the triangle OP1P2, in the ordinary sense of the term, is therefore equal to the absolute value of the expression l(xiy2 - XiVi). For some purposes it is convenient, however, to regard the area enclosed by a curve as a signed quantity, just as we have XI, § 195] THE STRAIGHT LINE 299 found it convenient to regard line-segments and angles as signed quantities. To this end, we observe that a point moving on the boundary of an area may make the circuit in either of two opposite directions (Fig. 170). With each of these directions is asso- ciated a definite rotation about a ^...,,^^ ^^^ point within the area. If the bound- ( ^^X ^ J^^ ary is traversed in a direction which \S^ X^x"^ produces a positive rotation about a positive arcuU Negative circuit point within the area, the circuit and ^^^- ^"^^ the area are regarded as positive ; if the boundary is traversed in the opposite direction, the circuit and the area are regarded as negative. Hence if an area is represented by a signed number, the sign of this number tells us the direction in which the boundary is traversed. In case of a triangle OP1P2 (Fig. 169) the order in which the vertices are written determines a direction of traversing the boundary. If OP1P2 is positive, OP2P1 is negative, and vice versa. Now in going around the triangle in the direction OP1P2, a segment OP joining to a point P moving on P1P2 generates a directed angle PiOP^- This angle is positive or negative according as the circuit OP1P2 is positive or negative. Moreover the measure of the angle PiOPo differs from 62 — ^1, if at all, only by multiples of 360°. The expression i OP, ' OP; sin((92 - ^1) is, therefore, positive or negative according as the circuit OP1P2 is positive or negative. We have then finally : The area of a triangle OP1P2 is given in magnitude and in sign by the formula (4) Area OP^P. = lixiUo - ^^2^1)- 300 MATHEMATICAL ANALYSIS [XI, § 196 196. The Area of Any Triangle. The convention as to the sign of an area is serviceable in deriving a formula for the area of any triangle in terms of the coordinates of its vertices. Let the vertices be Pi(xi, yi), P2(^) 2/2)) ^3(^*3? Ik)- Join these vertices to the origin by lines OPi, OP2, OPs. We novr con- sider the three possible cases, according as the origin is inside ><a! ^^ Fig. 171 '1^ Y X- \y X k Fig. 172 Fig. 173 (Fig. 171), outside (Fig. 172), or on, a side (Fig. 173) of the triangle PJ^JP^. Then in all cases, we have A P,P,P, = AOP.P, + AOP2P3 + AOP.Pi, if due regard is taken of the signs of the areas. Hence (5) Area of A P.P^P^^ \{if',x, - x^y, + y,x^ - x^y^ + y^x^ - x^y,). It might appear that this formula is difficult to apply. The following method makes it very simple. Write the coordinates of the vertices in two vertical columns as indicated, repeating the coordinates of the first vertex. Multiply each x by the y in the next row and add the products. This gives iCi?/2+X2y3+X3i/i. Then multiply each y by the x in the next row and add the products. This gives yiX2 + y^Xs + ysXi. Subtract the second sum from the first and divide the result by 2. The final result will be the area sought, with its proper sign-* A similar method may be used for finding the area of any convex polygon whose vertices are given. See Exs. 6, 7, 8, pp. 301, 302. * The student familiar with the elements of the theory of determinants will .observe that the area can be expressed as A=i Xi yi X2 Vi X3 ya Xi y\ Xl V\ 1 Z2 2/2 1 Xg Vs 1 XI, § 197] THE STRAIGHT LINE 301 Example. Find the area of the triangle whose vertices are Pi(— 4, 3), P2(— 1, - 2), PaC— 3, - 1). We write the coordinates of the vertices in two columns, repeating those of the first vertex. Performing the first step described in the previous paragraph ~ q ~ i we obtain 8 + 1 — 9 = 0; the second step gives — 8 + 6 + 4 = 7 ; the third step gives 0— 7 =— 7 ; dividing this by 2, we obtain — 3^ as the area of triangle P1P2P3. The magnitude of the area is 3^ square units, and the direction Pi to P2 to P3 is negative. Draw the figure and verify the latter statement. 197. Condition for CoUinearity of three Points. If three points Pi, P2, P3 are collinear, the area of the triangle formed by them is zero ; conversely, if the area of a triangle is zero, the three vertices are collinear. Therefore, a necessary and sufficient condition that three points be collinear, is that the right hand member of (5), p. 300 be equal to zero. EXERCISES 1. Find the areas of the following triangles and interpret the sign of the result in each case. Illustrate by appropriate figures. (a) (1, 3), (4, 2), (2, 5). (c) (- 5, 2), (- 4, - 3), (1, -1). (6) (2, 4), (- 3, 1), (1, - 7). (d) (a, a), (- 6, - 6,) (c, d). 2. Show that the following sets of three points are collinear : (a) (0,1), (2,5), (-1, -1). (c) (1,-2), (6, 1), (-4, -5). (&) (2, 1), (- 4, 4), (4, 0). (d) (0, -6), (1, a-6), (a, a^-b). 3. The point (h, h) is collinear with (2, 5) and (5, - 3). Find its coordinates. 4. Find the point on the y-axis collinear with (2, 5) and (5, — 3). 6. Under what conditions on a, b, c, and d are the points in Ex. 1 (d) collinear ? Interpret each of the conditions geometrically. 6. Area of any polygon. Show that the method of § 196 may be ex- tended to derive a formula for the area of any polygon in which two sides do not cross each other, and that if P1P2P3 -" Pn are the vertices of the polygon taken in order around the polygon, we have Area of polygon = A OP1P2 + A OP2P3 + A OP3P4 + ... + A OP„Pi, if due regard is paid to signs. 302 MATHEMATICAL ANALYSIS [XI, § 197 7. Find the area of the quadrilateral whose vertices are (1, 2), (-2,3), (-3, -4), and (4, -5). 8. Find the area of the polygon whose vertices are (4, 1), (2, 3), (0,4), (-2,3), (-4,1). 9. Prove that the points (1, 2), (3, 6), (—1, — 2) are collinear. 10. Show that the area of the triangle whose vertices are (2, 6), (— 4, 3), (—2, 7) is four times the area of the triangle formed by join- ing the middle points of the sides. 198. Applications to the Proof of Geometric Theorems. We shall now give a few elementary examples to show how the methods hitherto developed may be used in the proof of geometric theorems. Example 1. Prove that the line joining the vertex of any right tri- angle to the mid-point of the hypotenuse is equal to half the hypotenuse. Let ABC be any right triangle. In order to apply the methods of coordinates we must first locate a pair of coordinate axes. Any two perpendicular lines will serve the purpose, but the work incident to the solution of many problems may usually be greatly simplified if we choose the axes judiciously. In this case it is convenient (a,o')~x ^° choose the legs of the triangle as axes. The F 174. coordinates of the vertices are then (Fig. 174) (0, 0), (a, 0), and (0, &). The midpoint of the hypotenuse is (§ 194) (a/2, b/2). The length of the line joining this point to (0, 0) is V(a72JHjbj2y = ^VoM^ But the length of the hypotenuse is Va'^ + b'^. This proves the theorem. Example 2. Prove that the diagonals of a parallelogram bisect each other. Let ABCD be any parallelogram. Let a side of the parallelogram lie on the X-axis a vertex being at the origin. (See Fig. 176.) We may assign the coordinates {a, 0) to the vertex B, and (6, c) to the vertex D. The coordinates of C will then be (a-f-&, c). —q Why? We now calculate the coordinates of the •PCb.c) C(a+6.o) ^ £{0,0-) X Fig. 175 mid-point of ^C and also of the mid-point of BD, by the formula of § 194. It will then be seen that the midpoints coincide. XI, § 198] THE STRAIGHT LINE 303 Example 3. Prove that if the lines joining two of the vertices of a triangle to the mid-points of the opposite sides are equal, the triangle is isosceles. Let ABC be the triangle, M^ N the mid-points of the sides AC^ BC, respectively, with AN'= BM. Let the a;-axis lie along the side AB and let the y-axis pass through the vertex C (Fig. 176). Let the coordi- nates of A, B, C he (a, 0), (&, 0), (0, c) respectively.* We must first state the hypothesis of the theorem analytically, i.e. in terms of the coordinates. To this ^^' end we note that the mid-point of AC is M = {a/2, c/2), and that BM^' Y / CCo.c) J > P ^ X A(a,o) . B(b,o) Similarly, we have AN^ = (a-^Y-h- By hypothesis, AN = BM. Hence we have ■ This condition gives (-f)=.(«-|> which, when simplified, gives either a = b or a=— b. The first result would imply that the points A and B coincide, which is contrary to the hypothesis, and is therefore rejected. The second result yields readily that AC= BC, which was to be proved. EXERCISES 1. Prove analytically that the diagonals of a rectangle are equal. 2. Prove analytically that the line joining the mid-points of two sides of a triangle is half the third side. * In the figure a is a negative number. However, the discussion that follows applies at the outset to any numbers, a, b, c. It will appear later in the dis- cussion that, under the hypothesis of the theorem, a and b must have opposite signs. One of the advantages of the analytic method is the fact that it is general, and that ordinarily special cases do not have to be considered separately. 304 MATHEMATICAL ANALYSIS [XI, § 198 3. Prove analytically that two triangles with the same base and equal altitudes have the same area. 4. ABCD is a parallelogram, with A^ C as opposite vertices. JIf and iVare the mid-points of the sides AB and CD respectively. Prove ana- lytically that the lines AN and CM trisect the diagonal BD. 5. If P is any point in the plane of a rectangle, prove analytically that the sum of the squares of the distances from P to two opposite vertices of the rectangle is equal to the sum of the squares of the dis- tances from P to the other two vertices. 6. Prove analytically that, if the diagonals of a parallelogram are equal, the figure is a rectangle. 7. Prove analytically that the two straight lines which join the mid-points of the opposite sides of a quadrilateral bisect each other. 8. Show analytically that the figure formed by joining the middle points of the sides of any quadrilateral is a parallelogram. 9. If ilf is the mid-point of the side BC of any triangle ABC, prove that AB2+ AC^ = 2(AM^ + MC'^). 10. Prove analytically that the distance between the middle points of the non-parallel sides of a trapezoid is equal to half the sum of the parallel sides, 11. The difference of the squares of any two sides of a triangle is equal to the difference of the squares of their projections on the third side. 12. Prove that the sum of the squares of the sides of any quadrilateral is equal to the sum of the squares of the diagonals plus four times the square of the distance between the middle points of the diagonals. 13. If A^ jB, O, Z) are four points of a line prove the relation (due to Euler) : AB ■ CD-\-AC ■ DB-\-AD • BC=0. (The segments are directed.) 14. If M and iV, respectively, are the mid-points of two segments J.B and CD on the same line, show that 2 il/iV= AC + BD= AD + BC. 15. If 31 is the mid-point of AB and P any other point of the line AB, show that PAPB= PM^ - MAK 16. Two sources of light of intensity a and /3 are situated at the points A and B respectively of a line. Find the position of a point on the line which is lighted with the same intensity by the two points. How many points satisfy the relation ? [Hint : The intensity of light at a point varies inversely as the square of the distance of the point from the source of light and directly as the intensity of the source.] XI, § 198] THE STRAIGHT LINE 305 17. Two objects of weights Wi, w^ are situated at the points A^ Ao. The center of gravity of the two objects is defined to be the point of the line A1A2, whose simple ratio with respect to Ai, A2 is W2/W1. If Ai, A2 are on the x-axis, and their coordinates are Xi, X2^ find* the coordinate of the center of gravity. Show that the center of gravity does not exist, if Wi = — W2. Give an interpretation to a negative w. 18. Given n weights lOi, wo, ■•■■,Wn situated at the points Ai, A2, -••, A^ on a line. Find the center of gravity of ^1, A2 with weights Wi, W2 ; then the center of gravity of the point found taken with th^ weight Wi 4- W2 and A3 with the weight W3 ; then the center of gravity of this new point taken with "the weight Wi + W2 + 103 and A^ with the weight 104 ; and so on. Show that when all the ?i points have been used, there is obtained a point which is independent of the order in which the points were taken. The point thus determined is called the center of gravity of the n points. When does no center of gravity exist ? Under what coy- ditions is it indeterminate ? Show that if the latter conditions hold, each of the given points is the center of gravity of the remaining ones each taken with the weight assigned to it. 19. The first (or static) moment of a point P of weight w about a line I is defined to be the product of lo by the distance of P from I. Given n points Pi =CXi, yi)(i= 1, 2, .-., n) in a plane with weights lOc, respec- tively, determine the coordinates of a point P of weight lOi + «?2+ -•• 4- Wn such that its moment about the x-axis shall be equal to the sum of the moments about the x-axis of the points P,- and such that its moment about the y-axis shall be the sum of the moments about the y-axis of the points Pi. The point P is the center of gravity of the set of points. Com- pare with the result of Ex. 18. 20. The second moment or the moment of inertia of *a point P with respect to a line I is defined to be the product of the weight 10 of P by the square of its distance from the line. Given n points Pi in a plane whose distances from a fixed line I are Xi, and whose weights are Wi respectively. Let Ml be the sum of the first moments, M2 the sum of the second moments of these points about the line I. Let V be a second line, paral- lel to the first and h units from it (to the right or left according as h is positive or negative), and let Mi and iHf ^ be the sum of the first and sec- ond moments of the given points about I'. Let W be the sum of the weights Wi -{- 102 + ••• + Wn. Show that M'i = Mi-hW and M'2 = Mif-2hMi + h^W. 306 MATHEMATICAL ANALYSIS PCI, § 199 199. Directed Lines and Angles. An angle from a directed line li to a directed line I2 is an angle through which li must be rotated to . make its direction coincide with that of Zg. Any such angle we denote by {li y. Clearly if li and Zg intersect in a point M (Fig. 177), (Zi Z2) is the directed angle from Zi to l^ as ^^' ' defined in § 98 since the directions of Zj and Z2 define uniquely the half-lines issuing from M. As we observed in § 101, an angle (Zi I2) may have various determinations diifering from each other by multiples of 360°. The angle from the aj-axis to a directed line Z is called the inclination of Z (Fig. 178). If the inclination of a directed line Zi'is 61 and the inclination of a directed line l^ is $2, the angle from Zi to Z2 is given (§ 101) by the equation (6) (/i/2)=e2-ei, where the equality sign means equal except possibly for multiples of 360°. Fig. 178 200. Undirected Lines and Angles. If two lines Zi and Z2 are not directed, an angle from Zi to Z2, defined as an angle through which l^ must be rotated to make it parallel to Zg, will have various determinations which differ by multiples of 180° (Fig. 179). The smallest positive (or zero) angle from Zi to Z2 is then Fig. 179 unique and less than 180°. The inclination of an undirected line is defined as the smallest positive (or zero) angle through which it is necessary to rotate the a;-axis in order to make it parallel to the line. In Chapter III we used the slope m of a line to measure its inclination. It follows almost immediately from the definition of slope m and inclination 6 that we have m = tan 0. XI, § 201] THE STRAIGHT LINE 307 To calculate the angle from a line l^ to a line l^ we make use of (6), § 199, if the inclinations 0^, 6^ of l^, I2 are known. If the slopes mj, m^ of Zj, I2 are given, we lind from (6), § 138 tan (Zi /a) = tan (0. - O,) = tan 62 - tan 0^ ^ ' '^ ^ - /^ 1 + tan O2 tan ^1 But tan ^1 = wii and tan 0^ — ^2- Hence we have As special cases of this relation we obtain the familiar condi- tion for parallelism and perpendicularity (§§ 64, ^^^. For, if the lines are parallel, (Zi l^ — 0° or 180° ; hence m^ — m^. If the lines are perpendicular, (Zi, Zo) = 90° or 270° ; hence l-f-mim2 = 0, or mi= • mo 201. Standard Forms of the Equation of a Straight Line. We recall here for reference the standard forms of the equation of a straight line derived in Chapter III : The general equation : Ax -\- By -^ C = 0. The slope form : y = mx + b. The point-slope form : y — Ui = ^{x — Xi). The last two forms are not general, since they will not serve to represent lines parallel to the 2/-axis. The first is general. If the first represents a line not parallel to the ?/-axis (B =^ 0), it is readily reduced to the slope form, by solving the equation for y : A C y = X — — • ^ B B This yields, as was shown in § 63, A 308 MATHEMATICAL ANALYSIS [XI, § 201 EXERCISES 1. Construct a line through the point (— 2, 3) having an inclination of 00°. What is the slope ? Write the equation of the line. Find the points at which the line crosses the a:-axis and the y-axis. 2. Proceed as in Ex. 1 for a line passing through the point (2, — 3) with an inclination of 135°. 3. Find, to the nearest minute, the inclination of each of the follow- ing lines. Use a table of natural functions. (a) 2x-Sy = 0. (c) x = 2.1y + 3.5. (e) x - ?/ + 249 = 0. {b) y = OAx-{-l.'I. (d) 7x-\- Sy-8 = 0. (f)x + 2y+6 = 0. 4. Fhid the tangent of the angle from the first line to the second line of each of the following pairs. Then find the angle. {a)2x-Sy = 0, (c)x + 3y-3=0, ic + 2y+7=0. 3x — ?/ + 6=0. (b) 6x + 2y -10 = 0, (d) y = 2x + 3, 2x-|-32/ + 6 = 0. Sx + y-6 = 0. 5. Find the equation of the line through (4, 5) and parallel to the line joining (—1,2) and (2, — 3). 6. Find the equation of a line through the intersection of2x + y — 5=0 and x — 3y+5 = 0, and perpendicular to the line 2x — 3?/-fG = 0. 7. An isosceles triangle has for its base the line x—2 y+2=0 and for its vertex the point ( — 3, 5). The base angles are 45°. Find the equations of the other two sides and the coordinates of the other two vertices. 8. Given the lines aiX + biy + ci = and a^x + 62?/ -{- C2 = 0. Show that they are parallel, if and only if 05162 — «26i = ; and that they are perpendicular, if and only if aia2 + &162 = 0. 9. The sides of a triangle have slopes equal to ^, 1, and 2. Show that the triangle is isosceles. 10. Find the angles of the triangle whose vertices are (3, 4), (— 3, 6), and (2, - 1). 11. Find the slope of the bisector of the angle which a line of slope — 2 makes with a line of slope 3. 12. The slope of a line AB is 2. Find the equation of a line through the origin which makes with AB an angle whose tangent is — 1. 13. P is any point on the curve whose equation is 2/2 = 4 x. Show that the tangent to the curve at P bisects the angle which the line joining P to the point (1, 0) makes with the line through P and parallel to the x-axis. XI, § 202] THE STRAIGHT LINE 309 202. The Expression Axi-{-Byi-\- C. The expression x — 2y -f 3 has the value + 2 when x = l and y = 1', the value when x = —l and 2/ = 1 ; the value — 2 when a; = — 3 and y=l. The only interpretation we are able thus far to give to these facts is that the second set of values for x and y are the coordi- nates of a point (—1, 1) which is on the line whose equation is x-{-2y -{- 3 =0, while the other sets of values are the coordi- nates of points not on this line. It seems reasonable to expect, however, that the value of the expression xi—2yi-\-3, where (x^, y^) is any point in the plane, must have some relation to the line whose equation is x — 2y-\-3 =0. This relation is indeed very simple. * The reader should have no difficulty in proving that the value -f 2 obtained ^bove from the point (1, 1) represents in sign and magnitude the directed segment drawn parallel to the a;-axis from the line to the point (1, 1). Similarly, the value — 2 represents the segment drawn parallel to the a>axis from the line to the point (—3, 1). We proceed to show that a similar result applies to the values of the left-hand member of the equation of any line in the form Ax^By-\-C=0. Let the line I (Fig. 180) be the line whose equation is Ax-\-By-{- (7=0, where we assume A=^0, and suppose the equation has been written so that A is positive. Why is this last always possible? The line is then not parallel to the aj-axis. Why ? Let Pi{xi, yi) be any point in the plane and let Q(/i, y^ be the point in which the line through P parallel to the a;-axis meets I. Since Q is on ?, we have ^i , td , n a or By^-\-C = - Ah. e;^!^ Pi (x,.v,) Fig. 180 310 MATHEMATICAL ANALYSIS [XI, § 202 The value of Axi -{- Byi + C, wliicli we are seeking, is therefore equal to Axi — Ah, or A{xi — h). But Xi — h represents, in sign and magnitude, the segment QPi- We have then, Ax^ + By,+ C = A'QP;. We conclude that, HA is positive, the number Axi -f Byi + C is positive if {xi, 2/1) is to the right of the line Ax-\-By-{-C=0, and negative if (xi, 2/1) is to the left of this line. Moreover, Axi + Byi + C is proportional to the horizontal distance from the line to the point (a^i, 2/i)- Finally, if ^ = and B =^ 0, we may suppose the equa- tion By-{-C=0 so written that B is positive. The line I is then parallel to the a;-axis. Writing its equation in the form y =— C/B, it is readily seen that the expression -m 2/. -(t5^ 1 = 2/1+ J represents the directed segment drawn parallel to the y-asia from the line to the point Pi (Fig. 181). We may then con- clude that, B being positive, the number Byi-^-O ^li^vVi) is positive if the point (x^, y^) is above the line 1 By + C = 0, and negative if the point {x^, y^ is below this line. Moreover, Byi + C is jyro- portional to the distance of the point from the line. Fig. 181 ^ , ,. , ,. . . , By the preceding results, we may distinguish between the positive and negative sides of a line. If the equa- tion of a line is written in the form Ax -\- By -{- C = and so that its first term is positive, the right-hand side of the line is positive and the left-hand side is negative, unless the line is parallel to the aj-axis. In the latter case the upper side is positive and the lower side is negative. XI, § 203] THE STRAIGHT LINE 311 T -/; qA-^t,,., / / /^ X Fig. 182 203. The Distance of a Point from a Line. The results of tlie last article enable us to find the perpendicular distance of a point Pi(aJi, y^ from the line whose equation is Ax + By -f (7 = 0. If A 4=^, the required distance d = MPi (Eig. 182) is evidently equal to QPi sin^, where 6 is the inclination of the line. This is true whether the inclination is acute or obtuse, and whether P^ is on the positive or negative side of the given line. Since 0° ^ ^ < 180°, sin 6 is necessarily positive, and d = QP^ sin 6 will have the same sign as QPi ; i.e. it will be positive when Pi is on the positive side of the line, and nega- tive when Pi is on the negative side. We have, from the preceding article, QP _ Axi + By, + C and, since tan B = — A/B, we have sin^ = ^ VA^-{-& Hence, the required distance is (8) MPi=d = '^^'i±^Mi±£, If .d = 0, the required distance, by § 202, is simply , , C Byi + C ^^ B B But this is precisely what (8) becomes for ^ = 0. Hence (8) is true in every case. The distance d is positive if (xi, y^ is on the positive side of the line, and negative if (a^i, 2/1) is on the negative side, provided 312 MATHEMATICAL ANALYSIS [XI, § 203 the equation is written in the standard form with the first term positive. Example 1. To find the distance from the line 2a;— S?/— 10 =0 to the point (— 3, 1). Since the equation is in standard form the desired result is obtained by substituting the coordinates of the given point in the left-hand member of the equation and dividing by the square root of the sum of the squares of the coefficients of x and y. Hence the distance d ^® ^^ 2(-3)-5. 1-10 ^-21 V2'-^-|-(-5)2 V29 The negative sign indicates that the point is at the left of the line. Example 2. Find the equation of the bisector of the acute + 12 = and 4 ic - 3 ?/ + 6 = 0. First draw the lines (Fig. 183). know from geometry that the bisector of an angle is the locus of the points equidis- tant from the sides of the angle. Let (x, y) be any point on the desired bisector. In- spection of the figure shows that (aj, y) is on the positive side of one of the lines and on the negative side of the other. Hence, any point on the desired bisector must satisfy the condition that its distance from one of the lines is equal to minus its dis- tance from the other. This condition is expressed by the equation : between the lines 3 x -4y- p T7 A /. / /y y p' J ^^ / ^^'/ <-^J- 7 - ,' X -^i -f - -i^i 7 ■■ angle We Fio. 183 (9) or (10) 3 X — 4 j/ -f 12 5 4 a- - 3 y -h 6 5 18=0. 7x-7y Moreover, any point which satisfies relation (9) is a point of the bisector. Hence, we conclude that the equation 7x — 72/ + 18 = Ois the required equation. NoTK. Had the equation of the bisector of the obtuse angle been desired the figure shows that in this case a point on the bisector is either on the positive side of both lines or on the negative side of both lines. Hence, any such point must satisfy the relation obtained by placing its distance from one line equal to its distance from the other line. The equation of this bisector is x -f- y 4- 6 = 0. XI, § 203] THE STRAIGHT LINE 313 Example 3. Prove that the locus of a point which moves so that the algebraic sum of its distances from any number of fixed lines is constant, is a straight line. Each of the given straight lines has an equation of the form ax+by -\- c = 0. The distance of any point (x, y) from such a line is ax -\- by ■^- c ^ Va^ + &2 The equation of the required locus is, therefore, of the form aiX -hbiy -\- ci ^ ^ a nX + M + Cn _ q^ Since this is an equation of the first degree, the locus is a straight line. EXERCISES 1. Without using a figure determine whether the following points are at the right or the left of the line 2x + 3?/ - 5 = 0: (1, 2), (1, - 1), (- 2, 1),(1, 1), (4, - 2), (7, - 2), (4, - 1). Then, draw a figure con- taining the line and the points and verify the results obtained. 2. Find the distance of the point (3, —2) from the line 4x— 3y+6=0. 3. Find the distance of each of the following points from the line associated with it. In each case interpret the sign of the result. (a) (2,5),4x + 3y-2=0. (e) (- 4, 1), 3^/ - 2 = 0. (6) (-3, 7),5a; + 12?/+24=0. (/) (a, a), a: + ?/ - a = 0. (c) (2, - 2), 3 a: - 4 y = 0. {g) (6, a),ax + by = Q. {d) (5, 2),2x + 5 = 0. (Ji) (1,3), 2/ = 2 a; + 5. 4. Determine the region of the plane defined by each of the following sets of relations, (a) a; + 2 ?/ + 4 > 0, (&) 2 x - ?/ + 2 > 0, (c) 2 ic - 3 «/ + 6 > 0, a;_2i/-6>0. ?/-2<0. 3a; + 2y-12<0, x-y-l<0. 5. Define by inequalities (as in Ex. 4) the inside of the triangle ■whose sides are given by the expressions in Ex. 4, (c) equated to zero. 6. Define by means of inequalities the inside of the triangle whose vertices are (- 2, 5), (4, 1), (- 1, 1). 7. Find the distance between the two parallel lines 3x — 62/ + 5 = and 3x- 6?/- 2 = 0. 8. Find the equation of the bisector of the acute angle between the lines 2a; + 32/-4 = 0, x-2y + 7 = 0. 314 MATHEMATICAL ANALYSIS [XI, § 203 9. Find the equation of the bisector of the obtuse angle between the lines in Ex. 8. 10. Prove that the bisectors of the angles formed by the two lines a\X + 6iy + Ci = and a2X 4- h^y + c^ = are perpendicular to each other. 11. Find the lengths of the altitudes of the triangle whose vertices are (1,2), (-2,3), and (-3, -4). 12. Find the area of the triangle in Ex. 11 by multiplying half the length of one of the sides by the corresponding altitude, and check the result by finding the area by the formula of § 196. 13. Find the distance of the point (1, 2) from the line 3x + 4 ?/ + 12 = by finding the coordinates of the foot of the perpendicular dropped from the point on the line and then using the formula for the distance between two points. Check by means of § 203. 14. If the equations of two parallel lines are ax -\-hy + c z=Q and ax 4- by -f c' := 0, prove that the distance between them is the absolute value of (c — c')/y/a^ + h'^. 15. Prove that the bisectors of the angles of a triangle meet in a point. [Hint : Choose a convenient relation between the triangle and the axes. ] 16. Find the altitudes of the triangle formed by the lines x+2?/-3 = 0, x-y = 0, 4a;-y-l = 0. 17. Prove that the altitudes on the legs of an isosceles triangle are equal. 18. Prove that the three altitudes of an equilateral triangle are equal. 19. Prove that the sum of the absolute distances of any point within an equilateral triangle from the sides of the triangle is constant. 204. Two Equations representing the same Line. If of two equations of the first degree one can be obtained from the other by multiplying the latter by a constant, the equations obviously represent the same line, since all the points which satisfy one equation must then satisfy the other also. We now proceed to prove the converse of this statement : If the equations Ax + By + C = and A'x + B'y -{- C = represent the same line, either one can he obtained from the other by multiplication by a constant. XI, § 205] THE STRAIGHT LINE 315 Let us suppose first that none of the numbers A, A', B, B\ C, C is zero. The intercepts of the two lines on the ic-axis are then - C/A and - C'/A', on the y-Sixis - C/B and - C'/B'. Since the lines are by hypothesis identical, we have A A' ^. B B' From these relations follow at once A _B^_C__. A'~B~C'~ ' where A; is a constant. It follows that A^kA', B = kB', C=kC\ If C (or C") is zero, the corresponding line passes through the origin, and hence the other line must also pass through the origin ; hence C" (or C) is also zero. We leave the rest of the proof as an exercise, with the suggestion that the slopes of the two lines be compared. 205. The Intercept Form. Hesse's Normal Form. We have called attention thus far to three forms of the equation of a straight line : (1) the general equation ; (2) the slope form ; (3) the point-slope form. Two other forms are some- times of great convenience. These are the so-called mtercept form and normal form. The intercept form is (11) 1+1='^' ^"^^^'^ where a and h represent, respectively, the x- and ^/-intercepts of the line. This equation may be derived by finding the equation of the line through the points (a, 0) and (0, h). The derivation is left as an exercise. (See Ex. 21, p. 89.) This form is not applicable if the straight line passes through the origin, or if it is parallel to either axis. Why ? 316 MATHEMATICAL ANALYSIS [XI, § 205 The normal form is associated with the name of Hesse,* who used it extensively. It uses the length p of the perpendicular droj^ped from the origin upon the line and the angle a which this perpendicular makes with the a:-axis to determine the line. To derive the equation when p and a are given, we try to find a relation which is satisfied by the coordiDates (x, y) of every point P on the line and which is not satisfied by the coordinates of any other point. To this end (Fig. 184) we note that the projection of the broken line OMP on the perpendicular OQ is equal to p, if and only if P is on the line. The projections of the parts OM and MP on OQ are, respectively, x cos a and y sin a. The desired equation is, therefore, (12) X cos a + y sin a = ^ We shall take the positive direction of 0$, or p, from the origin towards the line, and choose the positive angle XOQ to be a. It is then evident that the position of any line is determined by a pair of values of p and a, it being understood that p and a are positive and that a is less than 360°. Moreover every line determines a single positive value oi p and a single positive angle a less than 360", unless p = 0. . When p = {) the line evi- dently passes through the origin and the above rule for the positive direction of p becomes meaningless. When p — 0, it is customary to choose a < 180°. To reduce the general equation Ax -{- By -\- C =0 to the normal form, we need merely observe that in the latter form an essential condition is that the coefficients of x and y are numbers the sum of whose squares is 1, since sin' a -|- cos^ « = 1. We must then multiply all the coefficients of Ax -{-By -^ C = by a number k, so chosen that (kAy -^(kBy = 1. This condi- ♦LuDWiG Otto Hesse (1811-1874), a noted German mathematician. XI, §205] THE STRAIGHT LINE 317 tion will be satisfied if ^ Therefore the desired reduction is obtained by dividing the equation through by ± V^^ + B^, and transposing the constant term to the right-hand side of the equation : A B ^ - (7 _ The sign of the radical must be chosen opposite to the sign of C, or if (7 = 0, the same as that of B. Why ? One advantage of the normal form is that every Hne may have its equa- tion written in the normal form. Whether the line passes through the origin or is parallel to an axis is immaterial. EXERCISES 1. Reduce the following equations to the normal form. Find in each case the values of a and p. (a) 4a; +8?/- 10 = 0. (d) 3x-2 y + 6 = 0. (6) a;-!/ + 5=0. (e)y=2x-3. (c) a; + VS ^ = 0. (/) a; = 2y - 5. (gr) The equation of the line whose intercepts are — 5 and 2, respectively. 2. Reduce to the intercept form each of the lines in Ex. 1 for which such reduction is possible. 3. What are the normal formsof the equations x =3,2 a:-f 3 = 0,^—1=0? 4. Derive the process of reducing the equation Ax + By -}- C = to the normal form by using the fact (derived from § 203) that p =— Cj^J A^-\-&. 5. What system of lines is obtained from the normal form, if a has a fixed value, while p is allowed to assume different values ? If p has a fixed value and a is allowed to assume different values ? 6. Find the equations of the lines which pass through the point (1, 2) and are two units distant from the origin. 7. Find the equations of the lines parallel to 5 x + 12 y = 13 and 3 units distance from it. 8. Find the equations of the lines parallel to 3 x + 4 ?/ = 13 and 7 units distance from it. 318 MATHEMATICAL ANALYSIS [XI, § 205 MISCELLANEOUS EXERCISES 1. Find the equation of the straight line passing through the point (3, 4), such that the segment of the line between the axes is bisected at that point. 2. Show that the lines y = ax -{■ a, for all values of a, pass through a fixed point, 3. Given aix + biy + ci = 0, aox + b2y + C2 = 0, agX + bsy + ca = 0, the equations of three lines forming a triangle. Show that the equation of any line Ax -^ By -^ C = in the plane may be written in the form ki{aix + biy + Ci) + koia-zx + b2y + C2) + ks^asx + bsy + ^3)= 0, where ki, k2, ks are constants. 4. Find the ratio in which the line Sy = 6 — x divides the segment joining the pohits (6, 1) and (— ^>, 2). 5. Find the equation of the line that passes through the point (1, 7) and makes an angle of 45° with the line x + 2 ?/ = 1. 6. Find the equation of the line that passes through the point (1, 7) and makes an angle of — 45° with the line x + 2y = 1. 7. Prove analytically that the perpendicular bisectors of the sides of a triangle meet in a point. 8. Prove analytically that the altitudes of a triangle meet in a point. 9. Prove analytically that the bisectors of the interior angles of a triangle meet in a point. 10. Prove analytically that the bisectors of two exterior angles of a triangle and of the third interior angle meet in a point. 11. Theequationsof two sides of a parallelogram are x— 2 i/=l, x+y=S. Find the equations of the other two sides if one vertex is at (0, — 1). 12. Find the equation of the line passing through the point (1, 1) and dividing the segment from (— 7, — 2) to (7, — 1) in the ratio 2:6. 13. Two vertices of an equilateral triangle are (1, 1) and (4, 1). Find the coordinates of the third vertex. There are two solutions. 14. jdind the equation of the line passing through the point (1, 2) and intersecting the line x -\- y = 4^ at a, distance ^VlO from this point. 15. Find the equation of the line through the point (1, 2) which forma the base of an isosceles triangle with the sides 2x — y = 1, x -{- y = I. 16. A straight line moves so that the sum of the reciprocals of its intercepts on the two axes is constant. Show that the line passes through a fixed point. XI, § 205] THE STRAIGHT LINE 319 17. If a straight line be such that the sum of the perpendiculars upon it from any number of fixed points is zero, show that it will pass through a fixed point. 18. Find the equations of the sides of the square of which two opposite vertices are (3, — 4) and (1, 1). 19. Derive the formula for the distance of a point (ici, ^i) from the line-^x i- By + C = hy finding the intersection of the perpendicular through the given point and the given line, and then using the formula for the distance between two points. 20. Prove that if the sum of the first moments of n points with respect to each of two given perpendicular lines is zero, the sum of the moments of tliese points with respect to any line in the plane through the inter- section of the given lines is zero. (See Ex. 19, p. 305.) [Hint : Take the given perpendicular lines to be the axes of coordinates. ] 21. If with the center of gravity of n points in a plane is associated the sum of the weights of the n points, prove that the sum of the first moments of the n points with respect to any line in the plane is equal to the first moment of the center of gravity with respect to the same line. 22. Given two half-lines r, s issuing from a point P, a third half-line t through P is completely determined if the ratio sin (r^)/sin (ts) = k is known. The ratio k is called the simple ratio of t with respect to r, s. Prove that the equations I = and m = of r and s, respectively, may be so written that, for all positions of t, the equation ot t is I — km = 0. 23. Given two points Pi(xi, y{) and P'z(x2, yi) and a straight line ax -\-hy + c = () which meets the line PiPi in Q. Find the simple ratio [Hint : This can be obtained very readily from a figure by observing the relation between the desired ratio and the ratio of the distances of Pi, Pi from the given line.] 24. From the last exercise derive the theorem of Menelaus : If a straight line cuts the sides of a triangle ABC in three points A\ B', C, the product of simple ratios AC BA> CB> C'B ' A'C' B'A is — 1. The point A' is on the side opposite A. B' on the side opposite JB, O on the side opposite C. CHAPTER XII THE CIRCLE 206. Review. The circle is the locus of a point which moves so that its distance from a fixed point, called the center, is con- stant. This constant distance is called the radius of the circle. If the center of a circle is at the point {h, k) and the radius is r, the equation of the circle is (1) {X - hy + (1/ - ft)2= r\ Tor, this equation expresses directly the fact that the square of the distance from the given point (7i, k) to the variable point {x, y) is r^. Hence, every point on the circle satisfies this equation and, conversely, any point not on the circle does not satisfy it. In particular, if the center is at the origin (h = 0, k = 0), the equation becomes (2) x' + y''^ 7-2^ "We note also that equation (1) when expanded has the form (3) x''-hy^ + Dx + Ey + C = 0. It follows that every circle in the plane may be represented by an equation of this form. To what extent is the converse true? Under what conditions does an equation of the form (3) represent a circle ? The answer to this question may be obtained by reference to the method of § 183. We desire to complete the square on the terms in x, and also on the terms in y. Therefore we rewrite the equation in the form ix^ + Dx-{- )-\.(y'^-^Ey+ ) = -C. 320 XII, § 207] THE CIRCLE 321 To complete the squares in the two parentheses we need to add i)Y4 to the first and J5jy4 to the second; to maintain the validity of the equation we must add the same terms to the right-hand member. We then obtain or (:H-f)V(..f)^=- + f--^. Since the sum of the squares of two real numbers is positive or zero, the left-hand member is positive or zero if x, y, D, E are real numbers. Hence the equation can be satisfied by real coordi- nates X, y only if L^ -\- E"^ — ^ C i^ di, positive number or zero. If Z>2+ ^ — 4 C is positive, equation (3) represents a circle with center at (— D/2, — E/2) and radius equal to I -y/D^ + E^-4.a If i)2_|_ ^2 _ 4 (7 ig zero, equation (3) is satisfied by the coor- dinates of the point (— -0/2, — E/2) and by the coordinates of no other (real) point. If Z)2 4- ^2 _ 4 (7 is negative, equation (3) represents no real locus. The answer to our question may then be formulated as follows : If (3) represents a curve at all, it represents a circle. 207. The Equation of a Circle satisfying given Conditions. The problem of finding the equation of a circle satisfying given conditions resolves itself simply into the problem of determining from the given conditions the values of U, Jc, r in equation (1), or of D, E, C in equation (3) of § 206. The fol- lowing examples will illustrate the methods that may be used : Example 1. Find the equation of the circle passing through the three points (3, — 5), (3, 1), and (4, 0). The desired equation must be of the form (8), and must be satisfied by Y 322 MATHEMATICAL ANALYSIS [XII, § 207 the coordinates of each of the three given points. If the first point satis- fies this equation, D, E, and C must be such that 32 +(- 5)-^ + ■i> • 3 + E(-6)+ C=0, i.e. such that SD-6E+ C=-S4. We find similarly from the second and third of the given points, 3i> + ^+O = -10, 4j[> + C=-16. Solving these three linear equations for Z>, U, C, we obtain Z)=-2, E = 4, C = -S. The desired equation is, therefore, x2 -J- ^2 _ 2 ic + 4 2/ - 8 = 0. Another method of solving this problem would be to regard (A, A) as unknown coordinates of the center. They must satisfy the two equations (3 _ A)2 + (_ 5 _ i.)2 =(3 _ hy +(1 _ j^y2^ (4-/i)2+(0- A:)2=(3-/02+(l-^)2. (Why?) By solving these equations we can determine h and k. Having found the center, it is easy to determine the radius. Then the desired equation can be written down in form (1). The completion of the work here sug- gested is left as an exercise. What other method could be used to solve this problem ? Example 2. Find the equation of the circle inscribed in the triangle ichose sides are y-S=0, 3x— 4?/— 9=0, and 12x+5y + 9 = 0. Let (h, k) be the center of the circle. It must be equidistant from the three sides. The distances of (A, k) from the three given lines are — (A: — 3), - ^ (3 A — 4 A: — 9), and ji^(12 ft + 5 ^• + 9), the signs being so chosen that each of these numbers is positive when (ft, k) is within the triangle. (See Fig. 185.) By placing the first of these distances equal to tlie second and third, re- spectively, we obtain two equations involving ft and k. The solution of these two equations yields ft = 1, k — \. Hence the center is the point (1, 1). The radius is evidently equal to 2. Why? Therefore the required equation is (a; - 1)2 +(2/ - 1)2 = 4, or a;2 + y2 _ 2x - 2t/ - 2 = 0. " X " --Y- ^ zzzzzY~.y-.-.iz HrnTNMTffl :i=::;z^ii=:=i: :: zt ::v= : :: Fig. 185 XII, § 207] THE CIRCLE 323 EXERCISES 1. Write the equations of the circles described below: (a) Center at the origin, radius equal to 5. (6) Center at (1, 2), radius = 4. (c) Center at (- 3, — 2), radius = 3. (d) Center at (a, a) and radius = a. (e) Center at (—2, 1) and passing through the point (3, — 2). (/) Center at (2, 1) and tangent to the x-axis. 2. Discuss fully the locus of each of the following equations : (a) x-^ + y^ - 2 X + 4:y -{- I = 0. {d) x'^+ y"^ -\- \ =0. (b) x^ + y^ -ix- 6y = 0. (e) x^ + y^ -\-2x- 6y + 10 = 0. (c) x^ -\-y^ + Sx-i = 0. (/) x2 4- 2/2 + 2 aa; + 2 a2 = 0. (g) Sx^ + Sy^-\-2x-iy-8=0. 3. What can be said of the coefficients Z>, E, and C in the general equation if the equation represents a circle which (a) passes through the origin ? (6) has its center on the x-axis ? on the y-axis ? (c) has its center on the line x -\-y = 0? (d) touches both axes ? (e) has its radius equal to 2 ? 4. Find the equations of the circles described below : (a) Passing through the points (0, 2), (1, 4), (1, 0). (6) Circumscribing the triangle whose sides are the lines «+ y — 3=0, x-2y + Q = 0, x + 2 = 0. (c) Inscribed in the triangle whose vertices are (0, 2), (0, — 4), and (-4,1). (d) Having ( — 2, 4) and (4, — 2) as the extremities of a diameter. (e) Passing through the points (1, 2) and (2, ]) and having its center on the line 2x + y + 2 = 0. (/) Tangent to both coordinate axes and passing through the point (2, 1). How many solutions are there ? 5. Prove analytically that any angle inscribed in a semicircle is a right angle. 6. Prove that the locus of a point which moves so that the sum of the squares of its distances from any number of fixed points is constant is a circle. Find the coordinates of the center of this circle in terms of the coordinates of the fixed points. If the number of fixed points is three, how is the center of the circle related to the triangle whose ver- tices are at the fixed points ? 324 MATHEMATICAL ANALYSIS [XII, § 207. 7. Find the equation of the locus of a point which moves so that the ratio of its distances from two fixed points is constant and equal to k. Determine fully this locus. Examine especially the case k=\. [Hint : Let the two fixed points be (a, 0) and (— a, 0)]. 8. Draw the loci of Ex. 7 for different values of k. Prove that if any one of these loci crosses the line joining the two given points in P and Q^ respectively, and the raid-point of the segment joining the given points is M, we have MP • MQ equal to the square of half the segment. 208. Tangent to a Circle. Point Form. In § 184 we saw how the slope of the curve Ax^ -f- Bi/ + Dx -\- JEy -{- C = at any point {xi, y-^ on the curve could be derived. Applying this method to the circle x'^j^if^Dx-\-Ey+ C = 0, we find the slope m at {x^, y-^ on the curve to be 2xy^ + D 2y,-\-E The equation of the tangent at the point {xi , y^ is, therefore, Simplifying, we obtain (4) 2x]X-\-2 y^y ^Dx-\-Ey -2 x^^ — 2y{'— Dx^ — Eyi = 0. But {xif z/i) is on the curve, and hence Xi'-hyi' + Dx, + Ey,-\-C=0. If this identity be multiplied by 2 and added to (4) we obtain 2x,x + 2y,y J^Dx-^Ey + Dx, -j- Ey^ + C = 0, or (5) x,x + y,y + 1 D{x + x,)-^l E{y + y{)-\-C=.0, As a special case of this equation (for D = 0, E = 0, C = — r^) we obtain the equation of the tangent to the circle x"^ -\- y^= 9-2 at the point (xi, y^) to be (6) x,x + y,y = r\ XII, § 209] THE CIRCLE 325 209. Tangent to a Circle. Slope Form. Another form of the equation of a tangent to the circle x^ -\- y"^ = r^ is often very serviceable. It is derived as follows. The straight line y = mx -\- h meets the circle x"^ -\- y"^ = r^ in points whose abscissas are given by the equation a;2 -I- {mx + hf = r». When expanded this equation becomes (1 + m2)a;2 -^2mhx -\- h"^ - r'^ = 0. The roots of this equation will be real and distinct, real and coincident, or imaginary, according as is positive, zero, or negative. Translated into geometric terms, this means that the line y = mx + h will meet the circle in two distinct points, two coincident points, or not at all, according as the expression above is positive, zero, or negative. If the line meets the circle in two coincident points, the line is a tangent. The condition ^ ^^^^ - 4(1 + m?) {¥ - r^) = yields, after simplification, or, b = ± rVl + m\ Hence, for all values of m the equation (7) y = mx ± r VI + m^ represents a tangent to the circle ^ -\-y'^ — r^. It follows at once that for all values of m the equation y-k= m(x -h)± r VI + m^ represents a tangent to the circle (x — hY-\-{y — ky = r^. 326 MATHEMATICAL ANALYSIS [XII, § 209 EXERCISES 1. Write the equations of the tangents to the following circles at the points indicated : (a) a;2 + ?/2 = 25, at (3, -4). (&) a;2 + y2 = 6, at(-l, 2). (c) a;2 + ?/2 = 4, at (0, 2). (d) jc2 + y2 _ 13^ at the points where x = 3. (e) a;2 4- ?/2 = 10, at the points where y = 1. (/) a;2 + 2/2 + 2 a: - 4 y = 0, at (1, 1). 2. Derive the equation of the tangent to the circle (x — h)^ + (y — k)^ = r2 at the point (xi, yi) by making use of the fact that the tangent is perpendicular to the radius through the point of contact. 3. Find the intersections of the following circles with the lines in- dicated : (a) x2-|-y2_ 5andy = 3a;-|-5. (c) ic2+?/2=:i3and3 a:+2 y-13=0. (6) x2 + r/2 = 25 andx - 2 !/ - 5 =0. (c?) x^ + 2/2 = 10 and y = 3 x + 10. (e) x2 -f ?/2 = 4, and y=-2x + 4, ?/ = -2x + 2 a/5, y = — 2 x + 5. Draw a careful figure showing the circle and the three lines. 4. Write the equations of the tangents to the following circles, the slopes of the tangents being as indicated. Find the points of contact. (a) x2 + 1/2 = 10, slope = - 3. {d) x2 + 2/2 ^ 25, slope = 0. (6)x2 + 2/2 = 5, slope = ^. (c) (x-l)2+(2/+2)2=10,slope=3. (c) x2 + 2/2 = 13, slope = |. 5. Will the equation y = mx ± rVl + m2 represent any tangent to the circle x2 + 2/2 = r2. Why ? 6. What is the point of contact of the tangent 2/ = wix + rVl + w2 to the circle x2 + y- — r^'i From this result derive the equation a^ia: + y\y = r2. 7. Any circle through the origin has an equation of the form x^+y^+Dx-{-Ey=0. Why? Prove that the equation of the tangent at the origin is Dx-^Ey=0. This may be done in at least two different ways. 8. Prove analytically that from an external point two real tangents can be drawn to a circle. 9. Derive the equation y = mx±r\/l -\- m^ directly from the property that a tangent to a circle is perpendicular to the radius through the point of contact. XII, §211] THE CIRCLE 327 210. The Value and Sign of the Expression x^ + y^ + Dx^ + Eyi + C. The left-hand member of the standard equation (x — hy +{y — ky = r^ represents the square of the distance from the point (x, y) to the point (h, k). Hence the expression (8) (^x^-hy+{y,-ky-r^ is positive, negative, or zero according as (xi, y{) is outside, in- side, or on the circle whose equation is (x — hy -\-{y — ky = r^. Moreover, from Fig. 186 it follows that if (xi, 2/1) is a point outside the circle, the expression (8) is equal to the square of the length ^ ^{x^,Vj) of a tangent drawn from the point (^'d 2/1) to t^® circle. Since the left- hand member of the general equation x^ -\- y"^ -\- Dx -{- Ey -\- C = may be writ- ten in the form (x — hy -\- {y — ky — r^ we may conclude that the sign of the -^ ^ Fig. 186 expression x^ -\- y^ + Dx^ -f- Ey^ + (7 is positive or negative according as the point (x^, y^ is outside or inside the circle x'^-{- y"^ -\- Dx -\- Ey -f C= ; and, if positive, it represents the square of the length of a tangent drawn from the point (xi, y{) to the circle. 211. The Equations of the Tangents from an External Point. Suppose we desire to find the equations of the tan- gents drawn from an external point {xi, yi) to the circle a;2 -f- 2/2 _ ^2^ Three methods will be discussed: Example, Find the equations of the tangents drawn from the point (4, - 3) to the circle x^ + y^ = 5. Method 1. Let {xi, y{) be the point of contact of one of the tangents. The equation of the tangent at this point is XiX -\- yiy = 6- However, since this tangent passes through the point (4, — 3) we have (9) 4 xi - 3 yi = 6. 328 MATHEMATICAL ANALYSIS [XII, § 211 But the point (cci, yi) is on the circle x^ -}- y2 _ 5^ Therefore ' (10) xi^ + yi'' = 5. Solving equations (9) and (10), we find the points of contact to be (2, 1) and ( — 2/5, — 11/5). Therefore the required tangents are2ic + y— 5 = and2a; + lly + 25 = 0. Method 2. From § 209 it follows that any tangent (not parallel to the ?/-axis) to the circle x^ -\- y^ = 6 is of the form y = mx ± VsVl + m^. Since this tangent is to pass through the point (4, — 3) we have — 3 = 4 m ± VoV 1 -j- W'^5 which simplifies to 11 m- + 24 m + 4 = ; th is giv es m =— 2, or — 2/11. Substituting these values in y — mx ± VSVl + m^ and simplifying we have 2x + ?/-5 = and 2 x + 11 y + 25 = 0. Method 3. The equation of any line through the point (4, — 3) is of the form ?/ + 3 = m(x — 4). Eliminating y between this equation and x2 + y2 — 5 ^e have (11) (wi2 + l)x2+ x(-8m2-6m) + (16m2 + 24m + 4)=0. Now since we desire y + 3 = w(:k — 4) to be tangent, equation (11) must have equal roots, i.e. (- 8 m2 - 6 m)2 _ 4(m2 + 1) (16 m2 + 24 m + 4) = or 11 m2 + 24 m + 4 = which gives m=— 2, or —2/11. Therefore the equations of the tangents are 2x + ?/ — 5 = and 2x+lly+25 = 0. 212. The Polar of a Point with respect to a Circle. Let us apply the first method of § 211 for finding the equations of the tangents from an external point to a circle, to the general problem of finding the equations of the tangent from the point (xi, 2/1) to the circle x^ -\- if- = r^. The coordinates {x\ y') of the point of contact are then found by solving simultaneously the pair of equations x'x^ + y'yi = r^, x'^ -f- y'^ = r\ The first equa- tion expresses the fact that the point (xi, y^ is on the tangent x^x -{- y'y = r"^ ', the second, that {x\ 2/') is on the circle. This shows that the straight line XiX + yiy = r^, where (xi, y{) is any external point, meets the circle in the points of contact of the tangents drawn from {xi, y-^. In other words, (12) iCiic + ;viy = 7-2 is the equation of the line joining the points of contact of the XII, § 212] THE CIRCLE 329 tangents through (xi, 2/1), if the latter point is outside the circle. If this point is on the circle, we know that (12) is the equation of the tangent at the given point. Finally, if (xi, y-^ is inside the circle, (12) represents a definite straight line determined by the point and the circle. This straight line (12), whether (^1) 2/1) is outside, on, or inside the circle, is called the polar of Fig. 187 (s^ij 2/1) with respect to the circle. The polar of (a^i, y^ with respect to a circle is then a uniquely determined line for every point (aji, 2/1) in the plane, except the center of the circle. (See Fig. 187.) Why this exception ? EXERCISES 1. Are the following points inside, outside, or on the circle x^ + y^ -2x + 6y-15=.0? (1,2), (1,0), (1,4), (-3,0), (3,0), (0,2), (5, 1). For the points outside, find the length of the tangents drawn to the circle. Draw carefully a figure to illustrate each of your results. 2. What is the length of the tangents drawn from (1, 1) to the circle whose equation m 2 r^ -\- '2, y'^ -{- ^ x — b y — \ = Q? [Caution : The equation is not in the standard form.] 3. Find the equations of the tangents drawn from the following points to the circle indicated : (a) (- 2, 4) ; a;2 + y2 = 10. (d) (3, 2) • x'^ + y^ = 4. (6) (5, _ 1) ; x2 + y2 = 13. (e) (4, 3) ; x2 + y2 = 16. (c) (3, - 1) ; x2 + 2,2= 2. (/) (7, 1) ; x2 + 2,2 ,, 25. 4. Find the equations of the tangents drawn from (0, 4) to the circle aj2 + y2 _ 2a; + 6y-15 = 0. 6. Show that the polar of a point P with respect to a circle is per- pendicular to the radius or radius extended through the point P. 330 MATHEMATICAL ANALYSIS [XII, § 212 6. Show that if P is inside the circle, the polar of P is wholly outside the circle. 7. Show that if the polar of P with respect to a circle whose center is cuts the line OP in Q, then OP • OQ=r^, where r is the radius of the circle. [Hint : Let the center O be the origin and the line OP the x-axis.] 8. Show that if the polar of a point with respect to a given circle is given, the point is uniquely determined. [Hint : This follows directly from the results of Exs. 5 and 7 ; or it may be proved directly by identifying the given polar ax-\-by + c = with the equation Xix + y\y = r^. In the latter case we should have Xi/a = pi/b = — r^/c, which determines xi, yi uniquely.] 9. A straight line is drawn through a given point P, cutting a given circle in the points A and B. Calculate the length of the segments PA and PB. Let P be chosen as origin and the line through P and the center of the circle as ic-axis. The equation of the circle is then x^ + y'^ 4- Dx 4- C = 0. If p is one of the segments PA or PB and « is the angle which PA makes with the x-axis, the coordinates of J. or P are {p cos a, p sin a). Since this point is on the circle we have the equation (/) cos a)2 + {p sin ay + !>/) cos « + O = for determining the two values of p. This equation reduces to p2 -f Z) cos a . /) + = 0. It may be noted that the product of the roots pip^ of this equation is C, i.e. independent of a. What theorem of elementary geometry does this" prove ? Prove also that the product PA • PB is positive or negative according as P is outside or inside the circle. 213. The Intersection of Two Circles. Given two circles 0^2 -f 2/2 4- D.a: + ^i2/ 4- Ci = 0, and ic2 4- 2/2 + D.p: + E^ + 0^= 0. The coordinates of the points of intersection are found by solving the equations simultaneously. Subtracting the equa- tions, we have (A-A)a5+(^i-^2)2/ + c,-a = o. Every point common to the two circles will satisfy this last equation, which is the equation of a straight line. Therefore the problem of finding the points of intersection of two circles XII, § 214] THE CIRCLE 331 is equivalent algebraically to that of finding the intersections of a straight line and a circle. This problem leads essentially to the solution of a quadratic equation in one unknown. There- fore we may conclude that two circles may intersect in two distinct points (two real roots), may be tangent to each other (coincident roots), or may not intersect at all (imaginary roots).* 214. Orthogonal Circles. Two circles which intersect at right angles are said to be orthogonal. In this case the tan- gents to the two circles at a point of intersection must be perpendicular, and the two tangents pass respectively through the centers of the circles (Fig. 188). The condition for orthogonality is then simply that the sum of the squares of the radii _ .„^ Fig. 188 of the circles shall be equal to the square of the distance between their centers. If the centers are (7i(7ii, ^i) and €2(712, k^ and the radii are Vi and r^ respectively, the condition for orthogonality is If the equations of the circles are ,. ox ^' + y' + ^^^ + ^^y + c'l = o» ^ ^ x^ + y'^ + D,x + Eiy+C2 = 0, this condition becomes (see § 206) 4 4 4 "^ 4 ' which when simplified gives A A + A^2 - 2(Ci + C2) = 0. * The reasoning above breaks down, if Z)i - 2)2 = and Ei — E2 = 0, that is when the circles are concentric. In this case, unless C\ —0^ = also (in which case the two circles coincide), the two equations are inconsistent and have no common solution, real or imaginary. 332 MATHEMATICAL ANALYSIS [XII, § 215 215. Pencil of Circles. Let the left-hand members of the equations (13), § 214, be represented by Mi and Mz respectively. Let us consider the locus of the equation (14) Mi-kM2 = 0, where k is an arbitrary constant. This equation may, iik^l, be written in the form (16) x^ + 2,^ + -L3^^» + -L_^^2/+-Y-r = 0. which represents a circle for each value of k{=^l). When k = l, equation (14) represents the straight line (16) (Z>i - D,)x + {El - E,)y + Ci - C^ = 0. The system of circles obtained by giving different values to A;, is called the pencil of circles determined by the two given circles. The straight line (16) is called the radical axis of the two given circles, and of the pencil. The following properties of a pencil of circles are readily proved : If the tivo gwen circles intersect in two points A and B, every circle of the pencil passes through A and B. If the two given circles are tangent to each other at a point Ay all the circles of the pencil are tangent at A. Tf trough any point in the plane not on the radical axis of the circles passes one and only one circle of the pencil. The proofs of these theorems are left as exercises. Further properties of pencils of circles will be found in the following exercises. EXERCISES 1. Find the coordinates of the points of intersection of the following pairs of circles : (o) x2 + y2 ^ 5 and a;2 + y2 ^ 2 X - 4 2/ + 1 = 0. (6) x2 + y2 _ a; _|. 2 J/ = and x^ + y2 4. 2 x — 4 y = 0. (c) x2 + y2 4. 2 X - 17 = and x2 + t/2 _ 13 _ q. XII, § 215] THE CIRCLE 333 2. Write the equation of the radical axis of each pair of circles given in Ex. 1. 3. Prove that the tangents drawn from any point of the radical axis of two circles to the two circles are equal. 4. Prove that the circles x^-{-y'^-\-6x — 2y + 2 = and x'^ +y'^ + 4y + 2 = are tangent to each other. Find their point of contact and the equation of their common tangent. 5. Find the equation of the circle through the intersections of the circles x^+y^ — 4x — 4 = and x^ + y^+2x — Qy — 2 = and the point (3, 3). [It is not necessary to find the intersections.] 6. Prove that the following circles are orthogonal: x^-\-y^ — 2x—4=0 and x^+y'^—Q ?/+4=0. In general for the circles : x^ -\- y'^ -\- Dx — C = and x^ + y"^ + Ey + C = 0. 7. Determine C so that x^ + y^ — 2x + 4:y — S = and x^ -\-y^ -\-2x + (7 = are orthogonal. 8. Prove that the locus of the centers of the circles of a pencil is a straight line perpendicular to the radical axis of the pencil. 9. Prove that if the radical axis of a pencil of circles is chosen as the y-axis and the line of centers as the ic-axis, the equation of any circle of the pencil is of the form x^ + y'^ + kx -\- C = 0, where C is the same for all circles of the pencil ; and that all circles obtained by varying k in this equation are circles of the same pencil. 10. The circles of the pencil in Ex. 9 intersect in distinct points, are tangent to each other, or do not intersect at all, according as C is negative, zero, or positive. In case (7 = 0, all the circles of the pencil are tangent to one another at the origin. Draw carefully three figures, illustrating the three kinds kinds of pencils here indicated. 11. Find the equation of a circle wbich is orthogonal to two given circles of the pencil in Ex. 9. [Hint : Let the two given circles be x"^ + y^ + kix + C=0 and x^ +y^ + kix + C = 0, and let the required circle be x"^ + y'^ + D^x + E^y + (^2 = 0. If this circle is to be orthogonal to each of the given circles we must have (§ 214) B^kx - 2((7 + (72) = and D^iki - 2(0 + (^2) = 0. These equations give 2)2 = and d^-C. Hence the required equation is x2+y2_|_^2?/— C=0. This yields two remarkable results : (1) The coeflB- cient E2 is undetermined, and by varying E^ we have a pencil of circles each of which satisfies the condition of being orthogonal to the two given 334 MATHEMATICAL ANALYSIS [XII, § 215 circles. (2) The equation found is independent of An, and k2- Hence, every circle of the pencil just found is orthogonal to each of the circles of the given pencil. Writing I for E2 to obtain uniformity of notation, wre have found two pencils of circles : and x2 + 2/2 4- A:x + O = «2 + 2/2 + ;y _ o = 0, such that every circle of either pencil is orthogonal to each circle of the other pencil. These two pencils of circles are said to form an orthogonal system. (See the adjacent figure.)] 12. In an orthogonal system of circles, the centers of the circles of one pencil are on the radical axis of the other pencil. 13. If the circles of one pencil of an orthogonal system intersect in two distinct points A and J5, the circles of the other system do not intersect at all, but pass between the points A and B. 14. If the circles of one pencil of an orthogonal system are mutually tangent to each other at a point A, the circles of the other pencil are also mutually tangent at A. 15. Prove that the three radical axes of three circles (not belonging to the same pencil) taken two by two intersect in a point. This point is called the radical center. Show that it is the center of a circle orthogonal to each of the three given circles and that the tangents drawn from it to the given circles are equal. XII, § 215] THE CIRCLE 335 MISCELLANEOUS EXERCISES 1. Find the condition that ax -{- by -\- c = be tangent to the circle «2 + y2 _ r2. 2. Find the equation of the circle passing through the points (0, 0), (a, 0), and (0, &). 3. Show that the equation of the circle having the points (a^i, yi) and (^2, 2^2) as the extremities of a diameter is (x— Xi)(x — a^a) + (.V — Vi) (y - 2/2) = 0. [Hint : The circle is the locus of the vertex of a right angle whose sides pass through the given points. ] 4. Find the equation of a circle w^hich is tangent to the lines a; = 0, y = 0, and ax + by -\- c = 0. 5. A line is drav^^n through each of the points (a, 0) and (—a, 0), the two lines forming a constant angle d. Find the equation of the locus of their point of intersection. 6. A straight line moves so that the sum of the perpendiculars drawn to it from two fixed points is constant. Show that it is always tangent to a fixed circle. 7. Give a geometrical construction for the polar of a point with respect to a circle. 8. If the polar of a point P passes through Q, then the polar of Q passes through P. 9. Find the equations of the common tangents of the circles x^+y^=S and x2 + ?/2 _ 10 ic + 20 = 0. 10. Find the locus of a point which moves so that the length of a tan- gent drawn from it to one given circle is k times the length of a tangent drawn from it to another given circle. 11. Find the equation of a circle through the points of intersection of 3:2 +2/2 _ 4 and x^-\-y^—2x-\-4:y+4:=0 and tangent to the line x—2y=0. 12. Show that the polars of a given point P with respect to the circles of a pencil pass through a fixed point, unless P is on the line of centers. 13. A point moves so that the sum of the squares of its distances from the sides of a given square is constant. Show that its locus is a circle. 14. A point P moves so that its distance from a fixed point A is always equal to k times its distance from another fixed point B. Show that its locus is a circle, if k =^1. Show also that for different values of k these circles have a common radical axis. 336 MATHEMATICAL ANALYSIS [XII, § 215 16. A line rotating about a fixed point meets a fixed line in a point P. Find the locus of a point Q on OP such that OP • OQ is constant. 16. Prove that among the circles of a pencil there are at most two which are tangent to a given straight line (unless all the circles are tangent to the line) . When is there only one ? None ? [Hint : Let the given line be the sc-axis.] 17. Inversion with Respect to a Circle. Given a circle with center and radius r. Corresponding to any point P in the plane (distinct from 0) there exists a unique point P' on OP such that OP • OP' = r^. The point P' is called the inverse of P with respect to the given circle. Prove the following propositions : (a) If P' is the inverse of P, P is the inverse of P'. (b) If P is inside the given circle, P' is outside ; and vice versa. (c) Every point on the given circle corresponds to itself. (d) If the coordinates of P and its inverse P' are (x, y) and (x', y') respectively, referred to two rectangular axes through 0, we have x'=-J^, y' = -'^; and x= ^^^' , y= ^^^^ • (e) If a point P describes a curve, the inverse P' describes a curve called the inverse of the former curve. The inverse of any straight line through is this line itself. (/ ) The inverse of any line not through is a circle through O, and the inverses of parallel lines are circles tangent at 0. (g) The inverse of any circle is a circle, unless the given circle passes through 0, in which case its inverse is a straight line. (h) Two orthogonal circles or lines have orthogonal inverses, (i) Any circle orthogonal to the given circle is its own inverse. (j) The adjoining figure illustrates a simple mechanism for changing circular motion into rectilinear motion. It is known as the inversor of Peaucellier. The heavy lines represent rigid bars, hinged at their extremities. The sides of the quadrilateral ABCD are all equal and OB - 0D= p. Prove that if O.is fixed and the mechanism is allowed to move in any way it can, C is always the inverse of A with respect to a circle with center O and radius r = \JP--p^, where I is the side of the rhombus ABCD. Hence, if A de- scribes a circle through 0, G will describe a straight line. CHAPTER XIII THE CONIC SECTIONS 216. Definition of a Conic. A conic section* or simply a conic is defined as the locus of a point which moves so that its distance from a fixed point, Fi, is always equal to a given constant, e, times its distance from a fixed line D^D^^. The fixed point F^ is called the focus. The fixed straight line A A' is called the directrix. The constant e is called the eccentricity. It is assumed that e > and that F^ does not lie on A A'. If P (Fig. 189) is any point on the curve, we have, by the preceding definition, (1) F,P = e' MP, where MP is the perpendicular distance of P from the directrix. It must be remembered that F^P and MP are absolute quantities, not directed quantities, and that e is positive. ♦ The name "conic section " is due to the fact that the curves in question were originally obtained as the sections of a right circular cone. They were discussed from this point of view by the ancient Greeks. z 337 338 MATHEMATICAL ANALYSIS [XUl, § 217 217. The Equation of a Conic. Let the directrix be chosen as the 2/-axis and the line through Fi perpendicular to the Y directrix as the aj-axis (Fig. 190). The coordi- nates of Fi may then be taken as (|), 0), y *' where p is different from zero. Let P (a;, y) be any point on the conic. Then A' P F^P.o)X F,P = ^{x-py + y% Fig. 190 ^^^ ^^ = + a; or - a; according as x is positive or negative. Equation (1), § 216 then becomes V(a;-i))2 + 2/2 = ±ea;. Squaring both sides of this equation and simplifying, we have (2) (1 - e2)a;2 + 3/^ - 2px-{-p'^ = 0. This is the equation of the conic. For, the coordinates of every point {x, y) satisfying the definition of the conic will satisfy equation (2), and conversely, every point whose coor- dinates satisfy equation (2) will satisfy equation (1). Why ? This is an equation of the type considered in § 183. It represents an ellipse if 1 — e^ > 0, a hyperbola if 1 — e' < 0, and a parabola if 1 — e'^ = 0. Hence we have, A conic is an ellipse, a parabola, or a hyperbola according as the eccentricity e is less than 1, equal to 1, or greater than 1. THE ELLIPSE 218. Standard Equation of the Ellipse : e < 1. We have seen in § 183 how to determine the locus of equation (2) by completing the square. If we apply the same method here, equation (2) may be written in the form (3) (1-e^-- XIII, § 218] THE ELLIPSE or (4) X — P T , 2/' _ pH^ ri _ p2\ ' 1 _ p2 ^1 — ^2\2 339 Since 1 — e^ is positive by hypothesis this equation represents an ellipse whose center is at the point {p/{l — e^), 0), and whose axes coincide with the two y\ straight lines x = p/(l — e^) and y = (Fig. 191). Let us move the curve parallel to the a;-axis through a distance -p/(l - e2), i.e. to the left if p > 0. Then its center comes to the origin, and its equation becomes Fig. 191 (5) or (6) x^ + p^e 2p2 1-e^ y2 (1 - e^y p^e^ p2^2 = 1. If we place (1 - e2)2 1 (7) (1-62)2 = «-, j9%2 = 6S the equation of the ellipse in its new position, i.e. with its center at the origin (Fig. 192), becomes Fig. 192 (I.) a2 &2 From (7) we have (8) 62=^2(1-62), which shows that b <a, since e < 1. If the ellipse is given in the form (IJ, a and b are known. Then the 340 MATHEMATICAL ANALYSIS [XIII, § 218 value of e can be found in terms of a and b bysolving equation (8) ; this gives (9) e^ = ^^A^ 219. Properties of the Ellipse. It is important to distin- guish between the properties of a curve as such and those properties which are concerned merely with the relations the curve bears to the coordinate axes. Thus the ellipse, as a certain kind of curve, is symmetrical with respect to two perpendicular lines called the axes of the curve. The longer of the segments on these lines cut off by the curve is called the majon axis, the shorter one, the minor axis. The inter- section of the two axes of the curve is called the center of the ellipse. Every ellipse, no matter how it is situated in the plane of coordinates, has a major axis and a minor axis as well as a center. From the way in which the equation was derived, we know also that every ellipse has a focus and a directrix. The symmetry of the curve with respect to the y-'dxis shows that this same curve could have been obtained from a second focus F2 and a second directrix D2D2 on the opposite side of the center. We shall now investigate how the two foci and the two directrices are related to the major axis, the minor axis, and the center. 220. Foci and Directrices. The original position of the focus Fi was (p, 0) ; the abscissa of its new position is ^ 1 _ e2 1 _ e2 Since from (7) we know that pe/{l — e^) = a, we find the coordinates of the focus F^ in the new position to be ( — ae, 0). XIII, § 221] THE ELLIPSE 341 (See Fig. 193.) Similarly the equation of the directrix AA' in its new position is X = P or (10) __a e The second focus F2 has the coordinates (ae, 0). The second directrix D2D2 has the equation FiQ. 193 (10') a X=~' e 221. The Ellipse in Other Positions. If the center of the ellipse is at the origin and the major axis is on the y-axis, the equation of the ellipse is (I.) 62 ^ a2 ' where, as before, 2 a is the length of the major axis and 2 6 is the length of the minor axis. The foci of this curve are at the points (0, ae), (0, — ae) ; the equations of the directrices are 2/ = ± « A- The equation of an ellipse whose center is at the point (h, k) and whose axes are parallel to the coodinate axes is (II.) (£^+(1^=1, (a>6) or (II„) .(l^ + (L=^^i, (a>b) according as the major axis is parallel to the a>-axis or to the y-Sixis. Finally we can reduce an equation of the form (III) Ax^-hBy^ + Dx-^Ey+C^^O, A>0,B>0, to the form II_, or 11^^, if it has a real locus. (See § 183.) 342 MATHEMATICAL ANALYSIS [XIII, § 222 222. The Case a = b. The Circle. If a = b the equation (Ij.) reduces to the equation of a circle. The relation a=b implies, however, that e = and this value of e is excluded in the definition of a conic. On the other hand it is clear that for a given value of a, as the eccentricity approaches zero, the ellipse approaches a circle. At the same time, the foci ap- proach the center, and the directrices recede indefinitely. Why ? Since the circle is a limiting form of an ellipse it is classified as an ellipse with equal axes and is counted among the conies. 223. A Geometric Property of an Ellipse. An important geometric property of any ellipse follows from the fact that the distance from the center to either focus, which we shall denote by c, is given by the relation or (11) c=:ae= V a^ ^S This relation shows that c, a, and h are the sides of a right- angled triangle in which a is the hypotenuse (Fig. 194). In other words, a circle drawn with its center at an extremity of the minor axis and with its 7'adius equal to a, will cut the major axis in the foci, Fi and F2. In computing the elements of an ellipse from a and b, it is generally con- venient first to find c from (11) and then to find e from the relation* (12) e = £ * This relation is equivalent to (9), § 218. It may be expressed by saying that e is the cosine of the angle CF^B, Fig. 194. XIII, § 224] THE ELLIPSE 343 The extremities of the major axis are called the vertices of the ellipse. The chord through a focus perpendicular to the major axis is called a latus rectum. Its length is 2 b^/a. Why ? _ ^ ^ -■■'^= W'' _ ^ s _, 5^^ I ^ Z .5 •3,22 * ^ 2 no) . _C:fi|Q) . J^Q) JC S (^ ^^=, --=^ rZ i ^ ±_ » : Fig. 195 224. Illustrative Examples. Example l. Given the ellipse 4a;2 + 9y2_36 = 0. Find the coordinates of the center, the vertices, the foci, and the equa- tions^ of the directrices. The given equation may be written in the form 9 4 from which follows that a = 3, 6 = 2. Therefore c = Va^ — b^ = \/5 and e = V5/3. The coordinates of the center are (0, 0), the vertices (3, 0) and (—3, 0), the foci (— V5, 0) and ( VS, 0) and the equations of the directrices are x = — 9/ V5 and x = 9/ VS (Fig. 195). Example 2. Find the coordinates of the center, the vertices, the foci, and the equations of the directrices of the ellipse 25 x* + 9 2/2 - 50 x+S6y - 164 = 0. From § 183, we know that the given equa- tion may be written in the form 25(x - 1)2 + 9(?/-f 2)2=225, or (x-1)^ , (y+2)2 ^.^ 9 25 We now conclude that the center is at (1, — 2), and that the major axis is parallel to the y-axis. Here a = 5, 6=3, c = 4, e = | Fig. 196 and a/e = ^^. Sketching the ellipse we find from the figure that the vertices are (1, 3) — 7), and the foci (1, 2) and (1, —6). The equations of the 33/4. ^ , " ~^-4 -'&" ^^ ^s^ ^ 0.2} ^ t X ^" (/r2) A ^ i \ "^^ ^^^a: -._ : ^± -: ::::i:i±li^:- and (1, directrices are y = 17/4, y 344 MATHEMATICAL ANALYSIS [XIII, § 224 EXERCISES In the following ellipses determine the major axis, the minor axis, the coordinates of the center, the coordinates of the vertices and foci, and the equations of the directrices. Sketch the curves. 1. 3 a-2 + 4 ?/2 = 12. 7. 3 a;2 + 3 2/2 = 12. 2. 4 a:2 + 3 ?/2 = 12. 8. x^ + 2y^= 8. 3. 4x2 + 2/2=16. 9. 4x2 + 9 2/2-16a;-18?/-23=0. 4. 36 x2 + 25 2/2 = 144. 10. 9 a;2 + 25 2/2 - 150 y = 0. 5. 2 x2 + 4 2/2 = 3. ^ 11. 4 a;2 + 2/^ - 8 a; + 4 2/ + 4 = 0. 6. 5x2 + 2/^^=75. 12. 9x2 + 42/2 + 36x-16 2/+16=0. 13. Write the equation of the following ellipses : (a) Center at origin, major axis = 4 on x-axis, minor axis = 3. (&) Center at origin, major axis = 5 on 2/-axis, minor axis = 3. (c) Center at origin, major axis = 6, minor axis = 3 (two solutions). (d) Center at origin, eccentricity 4/5, foci at (—2, 0) and (2, 0). (e) Center at (1, 2), major axis = 6 parallel to x-axis, minor axis = 4. (/) Foci at (0, 2) and (0, 8), major axis = 10. 14. An ellipse has its center at the origin, and its axes coincide with the coordinate axes. The ellipse passes through the points ( VT, 0) and (2, 1). Find its equation. [Hint. Assume the equation of the ellipse in the form (1^.). Find a and b from the fact that the ellipse must pass through the given points.] 15. Find the equation of the ellipse symmetrical with respect to the coordinate axes if the major axis is twice the minor axis and the curve passes through the point (2, 1). How many solutions ? 16. Show that the equation of the ellipse whose vertex is at the origin and whose major axis is on the x-axis is of the form a'^y- = ?>2(2 ax — :e2). 17. Verify equation (I^) by deriving the equation of a conic whose focus is at (— ae, 0) and whose directrix is the line x =— a/e. 18. Find the equation of the ellipse whose focus is at (0, 0), whose directrix is the line x + 2/ — 1 = and whose eccentricity is 1/2. 19. Find the equation of the ellipse whose eccentricity is 1/3, whose focus is at (3, 1) and whose directrix is the line 3x + 42/ — 1=0. 20. Find the equation of the conic whose focus is at (2, 1), whose eccentricity is 3, and whose directrix is the line 3 x + 2/ = 1. What kind of a conic is the curve ? XIII, § 225] THE ELLIPSE 345 225. Focal Radii. The segments F^P and F2P joining any point P on an ellipse to the foci JF\, F.^, are called the focal radii of the point P. If the equation of the ellipse is given in the standard form (/,), the focal radii of any poi7it P{xi, 2/1) «^e a — ea'i, a + exi. A Y P (x-„v,\ D, M, M, r> /K 4 \^i F.J X Fig. 197 For, from the definition of an ellipse (Fig. 197), • FiP=e'M,P, F.P=e'PM.^ But from the figure, we have also M,P='^ + x,, PM2 = --Xi. e Therefore the focal radii are FiP = a -f ex„ F2P —a — exi. From these relations follows the important property : The sum of the focal radii of any point of an ellipse is constant and is equal to the major axis 2 a. It may be noted that this relation still holds when the ellipse is a circle (e = 0), although the method of its derivation is not applicable in this case. An ellipse could, therefore, be defined as the locus of a point which moves so that the sum of its distances from two fixed points (the foci) is constant. 346 MATHEMATICAL ANALYSIS [XIII, § 226 226. Geometric Constructions of the Ellipse. The property of the ellipse derived in § 225 gives the construction indicated in Fig. 198 for the points of the ellipse when the foci and the major axis are given. 1 ^~r-^ Fig. 198 The segment AB is the major axis. Different positions of P on this segment give corresponding values AP and PB of the focal radii of a point on the ellipse. Circles drawn with these radii and centers at the foci intersect in points of the ellipse. To each position of P on AB correspond four points of the ellipse. A very convenient method of drawing an ellipse is indicated in Fig. 199. Two pins are stuck in the paper at the foci and Fia. 199 a loop of thread thrown over them. If a pencil point is in- serted in the loop and moved so as to keep the thread taut, it will describe an ellipse. Why ? Another method of constructing an ellipse (much used by draftsmen) is based on the fact (§ 179) that if the ordinates of the circle x"^ + 2/2 = a^ are shortened in the ratio b : a (b < a) XIII, § 226] THE ELLIPSE 347 there results an ellipse with major axis 2 a and minor axis 2 b. The adjoining figure (Fig. 200) exhibits the method. Explain and prove the method correct. Fig. 200 [Hint. The two circles being of radii b and a respectively, we have OB/OQ = b/a ; hence, MP/MQ = b/a. Why ?J EXERCISES 1. Construct an ellipse whose foci are 2 inches apart and whose major axis measures 3 inches. 2. Construct an ellipse whose major and minor axes are 2 and 1.6 inches respectively. 3. From the property of § 225 derive the equation of an ellipse. 4. From Fig. 200 show that the coordinates (x, y) of any point on the ellipse (Ij.), p. 339, are given bj'^ the equations x= acosQ, y = b sin 0, where 6 is the angle MOQ. Do these values of x, y satisfy the equation of the ellipse for all values of ^ ? 5. From the relation between the ordinates of a circle and an ellipse whose major axis is equal to the diameter of the circle prove that any plane section of a circular cylinder is an ellipse, provided the plane of section is not parallel to an element of the cylinder. 6. Prove from the result of the last exercise that a properly determined plane section of an elliptic cylinder is a circle. 348 MATHEMATICAL ANALYSIS [XIII, § 227 THE HYPERBOLA 227. Standard Equations of the Hyperbola. If e > 1, then 1 _ e2 < 0, and it is convenient to write (2), § 217, in the form (13) {f — V)x^ — 'f-\-2jix-f'^ 0. Completing the square and transforming as in § 218, we obtain / \2 2 ^^ra. This equation represents a hyperbola whose center is at the point ( — i)/(e^ — 1), 0) and whose axes coincide with the lines x = - j>l{f - 1), and 7/ = (Fig. 201). \:. a Xi \ 6 a >y A ^n y. /" K \v^\(fl^^o)*x /\ •>[ X Fig. 201 Fig. 202 If the curve is moved parallel to the a>-axis so that its center coincides with the origin (Fig. 202), its equation becomes y ^2g2 p^e^ = 1. (e2_i)2 e2_i If, then, we place the equation of the hyperbola becomes (Ix) XI II, § 2271 THE HYPERBOLA 349 ' Erom (14) we have the relation connecting a, b, e as 62 = a2(e2-l), or (15) e^ = ?i±*?. Here, as in the case of the ellipse, it is important to note some of the properties of the curve. It is seen that the locus is symmetrical with respect to the line passing through the focus and perpendicular to the directrix. This line is called the principal axis and the segment of this line intercepted by the curve is called the transverse axis and its length is 2 a. The extremities of the transverse axis are called the vertices, and the point midway between the vertices is called the center. The curve is also symmetrical with respect to the line through the center and perpendicular to the transverse axis. The seg- ment on this line whose length is 2 6 and whose mid-point is at the center of the hyperbola is called the conjugate axis. If a hyperbola has its center at the origin, and if its trans- verse axis 2 a is on the ?/-axis, and its conjugate axis is 2 b, its equation is (I) ^-y-=-i. The equation of a hyperbola whose center is at the point (h, k), whose transverse axis is 2 a, and whose conjugate axis is 2 b, is (IIJ (^_(J^ = 1, or (II.) (l|^-(l^^ = -l, according as the transverse axis is parallel to the x-axis or the y-axis. The equation of any hyperbola with axes parallel to the coordinate axes may be written in the form (III) Ax^-\-By^'-hDx + Ey-\-C=X), A>0,B<0; 350 MATHEMATICAL ANALYSIS [XIII, § 227 and every equation of this form (A > 0, B < 0) represents a hyperbola or a pair of straight lines (cf. § 183). As in the case of the ellipse, it is easy to show that every hyperbola has two foci on the transverse axis, one on each side of the center and at a distance c from the center, where c2 = 02^2 = a2 -I- bK With each focus is associated a directrix perpendicular to the trans- verse axis and at a distance a/e from the center (Fig. 202). The latus rectum, i.e. the chord through the focus and perpendicular to the transverse axis pro- longed, is of length 2 b^/a. The asymptotes of the hyperbola Fig. 202 (repeated) (x-hy {y-ky ^ ¥ = 1 are the lines (16) {^-hY &2 228. Geometric Properties of the Hyperbola. Tlie segment from the center to a focus of a hyperbola is the hypotenuse of a right-angled triangle ivhose legs are the semi-transverse and senii-coiij agate axes. Why? It is readily seen, moreover, that, if a rectangle be constructed by drawing lines through the extremities of each axis parallel to the other axis, the diagonals (extended) of this rectangle are the asymptotes of the hyperbola (Fig. 202). The circle drawn on either diagonal as a diameter passes through the foci. Why ? 229. Illustrative Examples. Example 1. Find the coordinates of the center, the vertices, and the foci, and the equations of the directrices and the asymptotes of the hyperbola 4x2-9.^2 + 36 = 0. XIII, § 229] THE HYPERBOLA 351 The equation is readily transformed into the form 9 4 It is now seen that the center is at the origin and that the transverse axis is along the y-axis (Fig. 203). The vertices are (0,2) and (0, -2). Since c = V^, the coordinates of the foci are (0, Vl3) and (0, — Vi3). The asymptotes are given by 4a;2_9y2_o or 2x -Sy = and 2x-\-Sy = 0. Since e = \/l3/2 the equa- tions of the directrices are y = ± Vis ±±vrs. ss^ F. (h, ^) >j/ i j^X'Z^t M s / T ^o^^ 1 ? D^ ^i--=^r ?i ^i^^ ^2^^N <^^ ^--'^ ^^^^ y^ t-Ta-^tS^ Fig. 203 Example 2. Find the coordinates of the center, the foci, and the ver- tices, and the equations of the asymptotes and the directrices of the hyperbola 16 x2 - 9 1/2 4- 32 X + 54 ?/ - 209 = 0. The given equation may be written in the form 16(x + l)2-9(y-3)2=144, or (a; +1)2 (i/-3)2 _.^ 9 10 The center is therefore at the point (—1, 3) and the transverse axis is parallel to the X-axis (Fig. 204). Since a = 3, the vertices are (2, 3) and (-4, 3). More- over, since c = V9 + 16 = 5, the foci are at the points (4, 3) and (- 6, 3). Like- 14 4 wise, e = c/a = 5/3 and hence the directrices are x = — — , x = - • The asymptotes are given by 16(a;+ 1)2 -9(2/ -3)2 = 0. Why? That is, the asymptotes are the lines 4x-Sy-\-lS = 0, and 4x-f-3y-5 = 0. Y \ ite "^ ^v 4^ _ Cv- :---= ^^ ^^^' ^^L ^t" M A ' /ji\ / \ '^ / \ ._S7_.X \ ~/^\ uJii- 3j::_ _vt/ ^Z ^ X2 X 7/^Z__::^g. " ^V ^ i^^^S Z^ M ^^ \^ 7 ^ Fig. 204 352 MATHEMATICAL ANALYSIS [XIII, § 229 EXERCISES For each of the following hyperbolas determine the transverse axis, the conjugate axis, the coordinates of the center, the coordinates of the ver- tices and the foci, and the equations of the directrices and asymptotes. Sketch the curves. 1. Sx^-4y^=12. 7. - 9 x2 + 2/'^ = 36. 2. 4 x2 - 3 2/2 = 12. 8. 2/2 - 2x2 = 4. 3. 4x2-3?/2z=_ 12. 9. 4x2- 12?/2_8x- 242/ - 56 = 0. 4. 3 x2 - 4 2/2 = - 12. 10. 5 x2 - 4 ?/2 -f 10 X + 25 = 0. 5. _36x2+25?/2 = 144. 11. 9x2 - 16i/2 + 18x - 96y - 279 = 0. 6. x2 - t/2 = 1. 12. x2 - 2/2 4. 2 X - 2 2/ = 2. 13. Write the equations of the following hyperbolas : (a) Center at origin, transverse axis = 6 on x-axis, conjugate axis = 4. (&) Center at origin, transverse axis = 8 on y-axis, conjugate axis = 10. (c) Center at origin, transverse axis and conjugate axis = 4, axes coinciding with coordinate axes. Two solutions. (d) Center at origin, focus at (5, 0) and transverse axis = 8. (e) Center at origin, transverse axis = 8, focus at (0, 5). (/) Center at Origin, focus at (5, 0), conjugate axis = 8. (g) Center at (1, 2), transverse axis = 6 parallel to x-axis, conjugate axis = 4. (A) Center at (0, 3), focus at (0, 5), conjugate axis = 2V3. (i) Foci at (1, 2) and (1, — 8), transverse axis = 6. 14. A hyperbola has its center at the origin and its axes on the coordinate axes; it passes through the points (0, VS) and (2, 3). Find its equation. [Hint. Since one point of the hyperbola lies on the ^/-axis, the equation may be assumed in the form I^^, i.e. 62 a2 and a and b may then be determined.] 15. Show that the equation of any hyperbola whose vertex is at the origin and whose transverse axis is on the x-axis is of the form a'^y- = hH2 ax -}- .x2). (See Ex. 16, p. 344.) 16. A hyperbola whose asymptotes are at right angles is called rectan- gular. Prove that the equation of a rectangular hyperbola may be written in the form x2 — w2 = cfi. XIII, § 231] THE HYPERBOLA 353 230. Focal Radii of the Hyperbola. If P{xi, y^) is any point on the hyperbola whose equation is i^2 _ ^ ^ the focal radii F^P and F2P are given by the equations F^P = exi + a, F^P = ex^ — a. The proof of the above statement is left as an exercise. It is analogous to the corresponding proof ^n the case of the ellipse (§ 225). Hence, the difference of the focal radii of any point on a hyper- bola is a constant. A hyperbola could, therefore, be defined as the locus of a point which moves so that the difference of its distances from two fixed points (the foci) remains constant. 231. Conjugate Hyperbolas. Any hyperbola determines uniquely a second hyperbola whose transverse and conjugate axes coincide in position and length with the conjugate and trans- verse axes respectively of the first hyperbola (Fig. 205) . Thus, if the equation of the first hyperbola is a2 62 1, the equation of the second hyper- bola is ^2 2/2 x^ a2 Fig. 205 Each of the two hyperbolas thus related is called the conjugate of the other, and the two hyperbolas are called conjugate hyperbolas. Two conjugate hyperbolas have the same asymptotes. Why ? 2a 354 MATHEMATICAL ANALYSIS [XIII, § 231 EXERCISES 1. Geometric construction of the hyperbola. Show how to construct a hyperbola given the foci and the length of the transverse axis by a method depending on the property of the hyperbola derived in §230 and entirely analogous to the first method described in § 226 for con- structing the ellipse. 2. Derive the equation of the hyperbola from the definition suggested at the end of § 230. [Let the foci be i^i(c, 0) and i^2(- c, 0) and let the constant difference of F\P and F2P be 2 a.] 3. What is the equation of the hyperbola x"^ — y^ — a^ after it has been rotated about the origin through an angle of 45° ? (Cf. § 190.) 4. From the result of Ex. 3 determine the length of the transverse axis of the hyperbola xy = k. 5. What are the equations of the hyperbolas conjugate to the hyper- bolas in Exs. l-12,p. 352? 6. Prove that the foci of two conjugate hyperbolas are on a circle. THE PARABOLA 232. Standard Equations of the Parabola. If in § 217 we let e = 1, equation (2) becomes (17) y^-2px+p^ = 0. or (18) y' = 'ip(^-fi- We saw in § 183 that this equation repre- sents a parabola whose vertex is at the point (p/2, 0) and whose axis coincides with the line y = (Fig. 206). If the curve is moved parallel to the ic-axis so that its vertex coincides with the origin, the equation of the curve becomes The focus of the curve is now at the point (i>/2, 0) and its directrix is the line x = — p/2 (Fig. 207). Xlir, § 233] THE PARABOLA 355 The following theorems follow directly. Their proofs are left as exercises. The equation of a parabola whose vertex is at the origin and whose axis coincides with the ?/-axis is (I,) x^=2/.j,. The equation of a parabola whose vertex is at the point {li, k) and whose axis is parallel to the a;-axis is (II,) (y-ky = 2p(x-h). Fj(±l\o) X Fia. 207 The equation of the parabola whose vertex is at the point (h, k) and whose axis is parallel to the 2/-axis, is (II,) {x-hy = 2piy-k). The equation of any parabola whose axis is parallel to the cc-axis is of the form (III.) By'^ + Dx-\-Ey^C = 0. The equation of any parabola whose axis is parallel to the 2/-axis is of the form (in J Ax^JrDx-\-Ey-\.C=0, The distance from the vertex to the focus and from the directrix to the vertex of the parabola y^ ==2px is p/2. 233. Geometric Properties of the Parabola. The chord drawn through the focus and perpendicular to the axis is called the latus rectum. Its length is twice the distance from the focus to the directrix. The focal radius connecting any point P{xi, 2/1) on the parabola y2= 2px to the focus is equal to Xi -{-p/2. The proofs of these properties are left as exercises. 356 MATHEMATICAL ANALYSIS [XIII, § 234 234. Illustrative Examples. Example l. Given the parabola r^ = 6 y. Find the coordinates of tlie vertex and the focus, and the equation of the directrix. Sketch the curve. The vertex is at (0, 0) and the axis of the curve coincides with the y-axis (Fig. 208). The distance from vertex to focus is 3/2. Therefore the focus is at (0, 3/2). Likewise, the distance from vertex to directrix is 3/2. Hence the equa- tion of the directrix is y =— 3/2. To sketch the curve, mark the focus, draw the latus rectum and then sketch the curve. 1 _:_ _ + ^_ \ : V ~-t L A -^ -"""*:- y ^ Fig. 208 Example 2. Given the parabola ?/2=:— 8x+2?/+15. Find the coordi- nates of the vertex and the focus, and the equation of the directrix. Sketch the curve. The given equation may be written as (y-l)2=-8(a:-2). Therefore the vertex is at (2, 1) (Fig. 209), and the axis is parallel to the a!;-axis. The distance from vertex to focus and from directrix to vertex is — 2. Therefore tlie focus is at (0, 1) and the equation of the directrix is x — 4t. The curve is readily sketched by plotting the focus and marking off the latus rectum. It may also be sketched by plotting another point or two. 1 'H ~ '*' s I s s \ (0, 1 J \: / / / / ^ T / X ' Fig. 209 EXERCISES Sketch each of the following parabolas. Determine the coordinates of the vertex and the focus, and the equation of the directrix. 1. y2 = 4a., 2. y^ =—4x. 3. 2/2 = 4 ic + 2. 4. y2__4a._^2. 5. x2 = 4 y. 6. x2=-4y. 10. a;2 -}- 4 X — 4 y + 6 = 0. 11. 2/2 _ 2a; -4?/- 8 = 0. 12. a:2 4- r/ -(- 1 = 0. 7. x^ = 4y +2. 8. x2 =-4?/ + 2. 9. 2/2 = 6x+ 12. 13. 2/2 = - 4 X + 2 y -f- 8. 14. y'^ + 2x-4y = 0. 16. x2 - 2 X -}- 2 y = 0. XIII, § 235] THE PARABOLA 357 16. Write the equation of each of the following parabolas : (a) Vertex at (0, 0) and focus at (2, 0). (6) Vertex at (0, 0), axis coinciding with ?/-axis, curve passing through the point (8, 4). (c) Focus at (— 1, 3) and directrix the line a; — 1 = 0. (d) Vertex at (1, — 2), axis parallel to x-axis, distance from vertex to focus equal to 2. (e) Vertex at (0, 2) , directrix parallel to a:-axis and parabola passing through the point (2, 1). 235. The Intersections of Conies and Straight Lines. The coordinates of the pomts of intersection of the ellipse (19) 6V -f ay = a'^b^ and the straight line (20) y = ma; + k, are found by solving these two equations simultaneously for (x, y). Eliminating y, we obtain the quadratic equation (21) {W + a}m')x^ + 2 a'mkx + a\k''- - h^) = 0, the roots of which are the abscissas of the points of intersection. For each of these roots the corresponding ordinate is found by substituting in (20). Why not in (19) ? We accordingly ob- tain, in general, two solutions {x, y). These solutions are real and distinct, real and equal, or imaginary, according as (22) 62'^ (j2^2 _ ]^2 ^0, =0, or < 0. Corresponding to these three cases, the straight line intersects the ellipse in two distinct points, in two coincident points {i.e. in a single point), or not at all. The discussion just given includes for a = 6 the case of the intersection of a circle and a straight line. To treat the intersection of the hyperbola ly^x^ — a V = «^^^ with the straight line (20), we need only notice that alge- braically we can reduce this problem to the preceding by simply writing — 6^ for h"^. Why ? 358 MATHEMATICAL ANALYSIS [XIII, § 235 This leads to the equation ((^2^2 _ 52)3.2 ^ 2 a^mkx + a2(fc2 + b"^) = 0. This is a quadratic equation unless a^mS _ 52 _ q^ if ^^2^2 _ 52 = 0, the line (20) is parallel to an asymptote, and, ii k ^ 0, it meets the hyperbola in only one point. If A: = the line is an asymptote and does not meet the curve at all. If a'^m'^ — b^z^O, we conclude that the line (20) intersects the hyperbola in two distinct points, two coincident points {i.e. in only one point), or not at all, according as (23) k'^-o?m^-\-¥>0, =0, < 0. Finally, the line (20) will meet the parabola (24) y' = 2px, in the points whose abscissas are the roots of the equation m2a;2 + 2(mk - p)x -\- k^ = 0. If m = 0, the line meets the curve in only one point. If m ^ 0, the line will intersect the parabola in two distinct points, two coincident points, or not at all, according as (25) p-2mk>0, =0, OY < 0. Similar results are evidently secured also for straight lines x = k, parallel to the 2^-axis. We then have the theorem : Any conic is met by a straight line in the plane of the conic in two distinct points, a single point, or not at all. EXERCISES 1. Draw figures illustrating all the results of the last article. 2. In a manner similar to that of the last article discuss the intersec- tions of the line y = mx -^ k and the conic y'^ = 2px-~ gx^. 3. Derive conditions analogous to (22), (23), and (25) of the last article when the straight line is assumed in the form Ax + By -\- C = 0. These conditions are slightly more general than those given in the text. Why ? XIII, § 236] THE PARABOLA 359 236. Tangents and Normals. Slope Forms. If, for a given value of m, the valae of k in the equation y = mx -f k is so deter- mined that the intersections of the line y = mx -\- k with a given conic coincide, i.e. so that the quadratic equation determining the abscissas of the points of intersection has equal roots, the line will be tangent to the conic. Why ? (See § 209.) The slope forms of the equations of the tangents to a conic result directly from the middle one of each of the conditions (22), (23), and (25) for the determination of k. Hence the equation of the tangent whose slope is m is : for the ellipse b'^x'^ -f a^y^ = a^b^^ (26) y =mx± Va^m^ + b^ for the hyperbola b'^x'^ — a^y^ = a^b"^, (27) y = mx± Va^m^ - b^ ; for the parabola y^ == 2 px, (28) y^mx+J--. Am We note that for a given slope the parabola has only one tangent, the ellipse two, and the hyperbola either two or none according as ahn^ — b^ '^ or < 0. [The condition a^m^ — &^ = yields the asymptotes.] The line drawn perpendicular to a tangent through its point of contact P is called the normal at P. EXERCISES 1. Find the equations of the tangents to the following conies satisfying the conditions given, and find for each tangent its point of contact : (a) 4x2+ 9y2 = 36, m = ^. (6) y^ = Sx, inclination 30°, 45°, 135°. (c) 9 x2 - 25 2/2 = 225, perpendicular to x + y -{-I =0. {d) x^ -y^ = 1, parallel to 5 x -f 3 y — 10 = 0. (e) y^ = 8x, perpendicular to 2x — Sy + Q = 0. 360 MATHEMATICAL ANALYSIS [XIII, § 236 2. Show that the line y = mx ± Vb'^ — a'^m^ is tangent to the hyper- bola &2a;2 — a2y'2 -f a2^2 _ Q for all real values of m lor which h^ — d^nf- > 0. 3. For what value of k will the line y =2x + k be tangent to the hyperbola x2 -4 2/2-4 = 0? 4. Find the coordinates of the points of intersection of the line 3x — w + l=0 and the ellipse x^ + 4 y2 = 55. 6. Find the points of contact of the tangents y = mx± Va^m^ + b'^ to the ellipse b'^x'^ + a^y^ = a^b'^. 6. From the result of Ex. 5 find the equations of the normals to the ellipse &2x- + a'^y'^ = a'^b'^ wliose slope is m. 7. By the method suggested in Exs. 5 and 6, find the equation of the normal to the hyperbola bH^ — a-y^ = a^b'^ in terms of its slope. 8. Same problem as Ex. 7 for the parabola y^ = 2 px. 9. A tangent to the ellipse b'^x:^ -\- a^y^ = a-b^ will pass through the point (xi, yi), if yi = mx\ ± Va'^m'^ + b'^. By solving this equation for m show that through a given point (xi, yi) will pass two distinct tangents, one tangent, or no tangents, according as b'^Xi^-^a^yi^—a^b'^ > 0, =0, or < 0. 10. By the method of Ex. 9, discuss the number of tangents that can be drawn from a given point (xi, yi) to the hyperbola b'^x^ — a^y^ = cfib'^', to the parabola y'^ =2 px. 11. Find the equations of the tangents to the parabola y^ = ix which pass through the point (—2, — 2). 12. Find the equations of the tangents to the ellipse 4 x^ + y2 = ig which pass through the points ( V3, 2) ; (0, 4) ; (0, 8) . 237. Tangents. Point Form. The slope of the curve (29) Ax^ + By^ + Dx + Ey-\-C= 0, at a point (a^i, 2/1) on the curve, was found in § 184 to be 2By,-\-E Hence the equation of the tangent to (29) at {xi, yi) is 2/ — 2/1 = — — -— ^ — (» — ^i)' This reduces to (30) 2 Axix + 2 Byiy + Dx-\-Ey = 2 Ax^'^ + 2 By^^ -f Dx^ -f Eyi XIII, § 237] THE PARABOLA 361 Since (xi, yi) is by hypothesis on the curve (29), we have 2 Ax,^ 4- 2 Byi^ = -2 Dx^-2 Eyi-2 C. Substituting this value in the right-hand member of (30), 2 Axx^ + 2 Byiy -\- Dx -\- Ey = - Dx^ - Ey^ -2 0. Hence, by transposing and dividing by 2, we obtain the equation of the tangent to (29) at the point (x^ y^ in the form (31) Ax,x + By,y + 2)(^^±^ + E^^^^ + C = 0. A A This equation is readily written down from (29) by replacing a;2, 2/2, X, and y by x^x., y^y, ^{x + x^), and }(y + y^), respectively. By applying this rule to the standard equations of the conies which are special cases of (29) we obtain : The equation of the tangent at the point (xi, y^ to the ellipse « y2 is Jfi^, l/i2/_1. a2 + b2-^' to the hyperbola aj2 y^ ^ a2 62 is to the parabola 2/2 = 2px is ViV = P{x + Xi EXERCISES 1. Write the equation of the tangent to each of the following conies at the point indicated : (a) x2 + 4i/2 = 8, at (2, 1). (6) 4a:2_3?/2 = 9, at (3, -3). (c) y'^ — Qx = 0, at the point where y = — 3. (d) x'^-y'^z=zA^ at (2, 0). (c) x2-2?/2^_4, at(-2, 2). (/) y2 — 4 X = 0, at the extremities of the latus rectum. 362 MATHEMATICAL ANALYSIS [XIII, § 237 2. Write the equation of the normal to each of the conies in Ex. 1 at the point indicated. 3. Find the equation of the normal to each of the conies hH"^ + ahj'^ = a^b^ b-x^ - aV = «^&^» and y^ = 2px at the point (xi, yi). 4. Prove that the tangents drawn to an ellipse at the extremities of any diameter (chord through the center) are parallel. 6. Show that an ellipse and a hyperbola with common foci intersect at right angles. 6. Show that the tangents at the vertices of a hyperbola meet the asymptotes in points at the same distance from the center as are the foci. 7. Find the angle (in degrees and minutes) at which the two curves a;2 + 2 y'-^ = 9 and ?/2 + 4 x = intersect. 8. Show that the secant of the parabola y^ = 2px joining the points (xi, yi) and (X2, yt) on the curve has the equation 2 px - (?/i + yi)y -\- yiy2=0. Show that this reduces to the equation of the tangent when the given points coincide. 238. Geometric Properties of Tangents and Normals to the Parabola. Let the parabola have the focus F, the vertex V, and the directrix d, the latter meeting the axis VF in D (Fig. 210). If the vertex is chosen as origin of a system of rectan- gular coordinates and the axis is chosen as the avaxis, while the segment DF is denoted by p, the equation of the parabola is y^ = 2 px. Now let P{x^, 2/0 be any point on the parabola. The equation of the tangent at this point is y^y =p{x -\-x^. This tangent meets the axis of the parabola (the a>-axis) in the point T{—x^y 0). Hence TV= VM, where M is the foot of the perpendicular dropped from P on the axis. From this, and by the definition of the parabola, d L V-^^^ /^ 'V \ T D Fig. 21 XIII, § 238] THE PARABOLA 363 follow the relations TF=DM=LP=FP, where L is the foot of the perpendicular drawn from P to the directrix. Hence TFPL is a rhombus. We conclude further that ZLPT=ZTPF', and, if S is the intersection of the diagonals of the rhombus TFPL, that the angle FSP is a right angle. Moreover, the line drawn through V, the mid-point of TM, perpendicular to TM, passes through S. We have then the following theorems : Theorem 1. Tlie tangent to a parabola at any point P bisects 07ie of the angles formed by the focal radius of P ayid the line through P parallel to the axis of the parabola ; the normal at P accordingly bisects the other angle. Theorem 2. The foot of the perpendicular dj-opped from the focus on any tangent to the parabola is on the tangent at the vertex. EXERCISES 1. Prove theorems 1 and 2 of § 238 analytically. 2. Give a geometric construction for the tangent to a given parabola at a given point. (The axis of the curve as well as the curve is supposed to be given. ) [A geometric construction means a construction with ruler and compass.] 3. Given the focus and directrix of a parabola, show how any num- ber of points of the parabola can be constructed on the basis of the results of the last article. 4. Given the focus of a parabola and the tangent at the vertex, use Theorem 2 of § 238 to draw any number of tangents to the parabola. These tangents will give a vivid picture of the shape of the curve ; the tangents are said to envelop the curve. The curve itself is not supposed to be given. 5. The outline and axis of a parabola are given ; show how to construct the focus and directrix. 364 MATHEMATICAL ANALYSIS [XIII, § 238 6. To construct the tangents to a giv6n parabola from a given external point. Assume that the focus and directrix and hence the axis are given. [Analysis: If Q is the given point, it follows from Theorem 1 of the last article that A QLiPi and A QFPi are congruent. Hence, Li ^'- ^ ^, kD KX l; 'XpF IV QLi = QF. We determine Zi (and Z2), therefore, as the intersection with the directrix of the circle with center Q and radius QF. Com- plete the construction. How is the con- struction affected when Q assumes various positions in the plane ? When is the construction impossible and why ? What happens when Q is on the curve ? 7. In the figure of Ex. 6, prove that the line through Q parallel to the axis bisects the "chord of contact" P1P2. 8. If a parabola is rotated about its axis the sur- face generated is called a paraboloid of revolution. Prove that if a source of light is placed at the focus of such a paraboloid*, all the rays issuing from the source will be reflected in the same direction (par- allel to the axis of the paraboloid). This is the prin- ciple of the so-called parabolic reflectors, used on searchlights, etc. 9. By an argument similar to that employed in § 212, prove that the equation of the chord of contact of the tangents drawn from an external point (xi, 2/1) to the parabola y'- = 2px is ijiy =p(x + Xi). This line is called the polar of the given point with respect to the parabola. It is defined by its equation even when no tangents can be drawn through the given point. 10. Prove that the polar of a point Q is parallel to the tangent at the point in which the line through Q parallel to the axis meets the parabola. 11. Prove that the length of the so-called subnormal My oi a parabola at the point P (see Fig. 210) is independent of the position of P on the curve. 12. Prove (Fig. 210) that TF = FN = FP and that FS=\ FN. ♦The focus of the generatmg parabolai is called the focus of the paraboloid. XIII, § 238] THE PARABOLA 365 13. Use the relation FN' = FP (Ex. 12) to show how to construct the normal at a given point P of a parabola (the focus and axis also being given). Construct a considerable number of normals in this way and show that they envelop a curve. (See Ex. 4 for the meaning of " envelop.") 14. Show that any two perpendicular tangents to a parabola intersect on the directrix. 239. Geometric Properties of Tangents and Normals to the Ellipse. If for any ellipse we let the coordinate axes coin- cide with the axes of the curve, the equation of the ellipse has the form The equation of the tan- gent at any point Pi(a;i, i/i) is b^x^x -h a%y = a^b^. _ A The a>-intercept (Fig. 211) of this tangent is Xi The remarkable thing about this result is the fact that it is independent of b and of y^. This means that if any other ellipse be given having the axis A' A in common with the first ellipse, then the tangent drawn to this new ellipse at a point having the abscissa x^ will also pass through T. This is therefore true of the circle drawn on A' A as diameter. If A' A is the major axis of the ellipse, this circle is called the major circle of the ellipse ; similarly the circle drawn on the minor axis of any ellipse as diameter is called the minor circle. * We do not in this article impose the restriction a > 6. 366 MATHEMATICAL ANALYSIS [XIII, § 239 A geometric construction for the tangent at any point Pj of an ellipse follows readily from the above considerations (as- suming that in addition to the curve one of the axes is given). Figure 211 shows the construction using the major circle and, in broken lines, the construction using the minor circle. The following theorem is of fundamental importance in dis- cussing the geometric properties of the ellipse : Theorem 1. Tlie tangent and the normal to an ellipse at a given point bisect the angles formed by the focal radii drawn to the point. Proof. We are to prove that the tangent at Pj (Fig. 212) bisects the angle F^PxR, and that the normal at P^ bisects the angle F^P^F^. To this end we calculate first the tangent of the angle SP^R. Using the equation of the ellipse as given above and taking the foci to be PgCc, 0) and Pi(— c, 0), we have the slope of the tangent P^S = - ^, a^y^ the slope of F,R (i.e. FiP,) = -^ — The tangent of the angle <^i from P^S to P^R is then Xi -\- c a^yi tan d)i — — ^' 1 b%y, a%{x, + c) Simplifying this expression, we find tan <^i = — -. The tangent of the angle <^2 from P^S to P^F^ may evidently be obtained by simply changing c to — c in the last result. (Why?) Hence, ., tan <f)2== • XIII, § 239] THE PARABOLA 367 We conclude that <^2 = — ^v This proves that PiS bisects the angle F^PiB. That the normal bisects the angle FiP^F^ follows at once from elementary geometry. The theorem just proved leads at once to another geometric construction for the tangent (and normal) to an ellipse at a given point, supposing the foci of the ellipse are known. Theorem 2. The foot of the perpendicular dropped from either focus on any tangent to an ellipse lies on the major circle. Proof. (See Fig. 212.) Let S be the foot of the per- pendicular dropped from F^ on the tangent P^S, and let it meet the line F,P^ in R. Then F^^P^B is an isosceles triangle (why?) with P^R = P.F^. We have then F,R = F,P, + P,F^ = 2 a. (§ 225) Also aS' is the mid-point of F2R and O is the mid-point of FiF^. Hence OS=^FiR==a, and S is on the major circle. We should note also that, if Q is any point on the tangent PiS, then QR = QF2, which is important in connection with the problem of drawing the tangents to an ellipse from an external point. (See Ex. 5, below.) Fig. 212 EXERCISES 1. Show how to construct the tangent to a given ellipse at a given point. (Two constructions, one using the major circle, one using the foci.) 2. Show that, in Fig. 211, OA is a mean proportional between OM and OT. 3. Show that, in Fig. 212, OFi is the mean proportional between the intercepts on the cc-axis of the tangent and normal at Pi. 368 MATHEMATICAL ANALYSIS [XIII, § 239 4. Prove analytically that S (Fig. 212) is on the major circle. 5. Show how to construct the tangents to an ellipse from a given ex- ternal point Q. [Hint : Construct B (Fig. 212) as the intersection of two circles, one with center Fi the other with center Q.] 6. Show that if a right angle moves with its vertex on a given circle and one of its sides passing through a fixed point within the circle, the v-^ther side will envelop an ellipse. 7. Use the result of Ex. 6 to construct a considerable number of tan- gents to an ellipse, given the major circle and one focus (the outline of the ellipse is not supposed to be given in advance, but will appear vividly after this problem is solved). 8. If an ellipse is rotated about its major axis the surface generated is called a prolate spheroid. Show that sound waves issuing from one focus will be reflected by the surface to the other focus. This principle is used in the so-called ' ' whispering galleries. ' ' 9. By an argument similar to that used in § 212 show that the equation xxi/a^ -f yyi/b'^ = 1 is the equation of the line joining the points of contact of tangents drawn from (xi, yi) to the ellipse x^/a''^ + y'^jW- — 1. 240. Geometric Properties of the Hyperbola. Many of the geometric properties of the hyperbola are similar to cor- responding properties of the ellipse, which is to be expected in view of the similarity of their equations. The following two theorems are fundamental : Theorem 1. Tlie tangent at any point of a hyperbola bisects the angle betiveen the focal radii di-aion to the point. The normal bisects the adjacent supplementary angle. Theorem 2. The foot of the perpendicular dropped froin either focus on any tangent to a hyperbola is on the circle drawn on the transverse axis as diameter. The proofs of these theorems are left as exercises. They are similar to the proofs of the corresponding theorems on the ellipse. Draw figures illustrating Theorems 1 and 2. XIII, § 240] THE PARABOLA 369 Certain new properties of the hyperbola relating to the asymptotes will be found among the exercises below.. EXERCISES 1. Show how to construct the tangent and the normal to a given hyperbola at a given point. 2. If P is any point on a hyperbola, OA the semi- transverse axis, ifcf the foot of the perpendicular dropped from P on OA (produced), and T the point in which the tangent at P meets OA, prove that OA is a mean proportional between Oilf and OT. 3. With the notation of Ex. 2 show that OFi is the mean propor- tional between ON and OT, Fx being the focus on OA and N the point in which the normal at P meets OA (produced). 4. Prove Theorem 2 (§ 240) analytically. 5. Show how to construct the tangents to, a hyperbola from an ex- ternal point. 6. Show that if a right angle moves with its vertex on a given circle and one of its sides passing through a fixed point outside the circle the other side will envelop a hyperbola. 7. The construction of tangents to a hyperbola analogous to Ex. 7, p. 368. 8. Use Ex. 3 above and Ex. 3, p. 367, to show that an ellipse and hyperbola having the same foci intersect at right angles , 9. Prove that, if a tangent to a hyperbola meets the asymptotes in Ti and T2, the point of contact of the tangent is the mid-point of the segment T1T2. 10. Prove that the area of the triangle formed by any tangent and the asymptotes of a hyperbola is constant (= a6). 11. Show that if a straight line cuts a hyperbola in Pi and P2 and the asymptotes in ^1 and ^2 the segments PxQx and P2^2 are equal. Use this result to construct any number of points of a hyperbola when the asymptotes and one point of the curve are given. 12. By an argument similar to that used in § 212 show that the equation xxx/a^ — yyi/h^ = 1 is the equation of the line joining the points of contact of the tangents drawn from (cci, 2^1) to the hyperbola a;2/a2 - 2/2/62 = 1. 2b 370 MATHEMATICAL ANALYSIS [XIII, § 241 241. The Conies as Plane Sections of a Cone. We stated in § 216 that the ellipse, hyperbola, and parabola could all be obtained as the plane sections of a right circular cone. This we shall now proceed to prove. In doing so we shall get the machinery for solving problems of a more general type. If a point P in a plane a (Fig. 213) is joined to a point S not in a by a straight line SP, the intersection P' of SP by a plane a' is called the projection of P from S upon a'. Similarly, if all the points of a curve in a be joined to S, the intersections of these lines with a plane a' form a curve C", Fig. 213 ^^v/ which is called the projection from S of the curve C. The point JS is called the center of projection, and the process described is called central projection, to dis- tinguish it from orthogonal projection previously considered {e.g. in § 135). If, now, the curve C in the plane a is a circle, the lines through jS and the points of this circle form a cone with vertex S. This is not a right cone, in general. As the lines through S are not supposed to terminate in S, we get a so-called co7n- plete cone, or cone of two nappes, which consists of two con- gruent ordinary cones placed vertex to vertex so that their axes form a straight line. It will now be clear that a plane section of this cone is the same as the projection of the circle C from the vertex S upon the plane of section. We have then reduced the problem to that of finding the central projection of a circle. We will solve it by finding the relation between the coordinates of a point P in a and the co- ordinates of the corresponding point P' in a'. To this end (Fig. 214) let be the foot of the perpendicular dropped from JS on the line of intersection of the planes a and a'. Let be XIII, § 241] THE PARABOLA 371 the origin and let the line of intersection OF of the two planes be the 2/-axis in the system of coordinates in each of the two planes. Let the line OX perpendicular to 01^ in a be the a^axis in a, and the line OX' perpendicular to OF in a' be the ic-axis in a'. Let the angle between the two planes be ; then X'OX= 0. Now let P{x, y) be any point in the plane a, and let P\x\ y') be the projection of P from S. We seek the relation connecting the coordinates x, y, x', y'. Draw ST parallel to OX, and represent the length OS by h. Then we have T0=^, TS= ^ sinO tand We then have from similar triangles x' :{TO-i-x') =x:TS, y':y = SM^ : SM= TM' : TO. If we substitute the values of TO, TS, and TM' (= TO + x'), we obtain , x' + h h ' sin tan 372 MATHEMATICAL ANALYSIS [XIII, § 241 and , , h sind 1^ y _Jl_ sin B Solving these equations for x and ?/, respectively, we have _ li cos 6 • x' _ hy' ^~ ^mO'X' + h' ^"sin^-x' + Zi* If these expressions be substituted for x and y in the equa- tion of any curve in the plane a, the resulting equation in x' and y' will be the equation of the projection of the curve in a'. To solve the problem we proposed at the outset, let the curve in the plane a be the circle x'^-\-y^=a\ The equation of the corresponding curve in a' is then 7^2 cos2 ^ . x'2 -f hhj"^ = tt2 sin2 e-x'^+2 lia^ sin • x' -\- a%\ Collecting like terms, we have * (/i2 cos2 e-a" sin2 6) x"- + li'y'^ - 2 ha" sin 6 - x' - aVi^ = 0. We see at once that this is the equation of a conic. It is an ellipse, a parabola, or a hyperbola according as /i2cos2^-a2sin2^>0, = 0, or < 0, i.e. according as tan ^ - - < 0, = 0, or > 0. a But h/a is the tangent of the angle <^ which an element of the cone with vertex S makes with the plane a. If is less than this angle (f>, the section of the cone is an ellipse ; if ^ is equal to <^, the section is a parabola ; and if 6 is greater than <^, the section is a hyperbola. Note that this result is in accord- ance with our geometric intuition of the situation. XIII, § 242] THE PARABOLA 373 EXERCISES 1. Prove that the central projection of any circle is a conic ; that is, that a plane section (not through the vertex) of any circular cone (not necessarily a right cone) is a conic. [Hint : Complete generality will be secured by taking the equation of the circle in a to be x'^ ^- y'^ -{■ dx -\- c = 0. Why ?] 2. Prove that the central projection of any conic is a conic. ,3. Prove that the central projection of a straight line is a straight line. 4. Prove that there exists in a just one straight line which has no corresponding line in a', namely the line of intersection of a with the plane through S parallel to a'. This line is called the vanishing line of a. 5. Prove that the central projection of a circle in a is an ellipse, a parabola, or a hyperbola, according as the vanishing line in a meets the circle in no points, one point, or two points. 242. Poles and Polars. Diameters. We have had occasion in several exercises to note that the equation which represents the tangent to a conic at the point Piix^ , y^ when Pj is on the curve, represents a straight line called the polar of Pi when Pj is any point in the plane. Pi is then called the pole of the line with respect to the conic. The polar of a point on the conic is then the tangent at the point. We have also seen that the polar of a point Pi through which pass two tangents to the conic is the line joining the two points of contact of the tangents. In the more extensive geometric theory of conies poles and polars play an important role. A straight line passing through the center of an ellipse or hyperbola is called a diameter of the conic. Every diameter of an ellipse meets the curve in two points ; some of the diameters of a hyperbola meet the curve in two points. These points are then called the extremities of the diameter, and the distance between them is called the length of the diameter. Any line parallel to the axis of a parabola is called a diameter of the parabola. Other properties are given in exercises below. 374 MATHEMATICAL ANALYSIS [XIII, § 242 MISCELLANEOUS EXERCISES Properties of Poles and Polars 1. Write the equation of the polar of each of the following points with respect to the conic given, and draw the corresponding figure : (a) (1,5); 2a;2 + y2^4. (d) (- 1, 3) ; a;2 + 2/2 + 4 x - 6 y- 2 = 0. (6) (2,0); 4x2-92/2 = 36. (c) (2, - 3) ; 2/2 = 6x. 2. Find the pole of the line 3x — 4y4-12 = with respect to tho following conies : (a) 3x2 + 4 y2 = 12 ; (ft) a;2 - 5 y2 = 20 ; (c) y^ = Sx; (d) x'^ = iy. 3. Prove that in any conic the polar of a focus is the corresponding directrix. 4. Prove that in any conic, if Pi and P2 are two points such that the polar of Pi passes through P2, the polar of P2 will pass through Pi. 5. From the result of the last exercise follows geometrically the follow- ing theorem : If a straight line be revolved about a point P and tangents are drawn at the points where it meets a conic, the locus of the intersec- tion of these pairs of tangents is the polar of P with respect to the conic, or a part of the polar. Which part will it be ? 6. Prove that the polar of any point on a directrix of a conic passes through the coiTesponding focus. [See Exs. 3 and 4.] 7. A straight line through a point Pi meets a conic in Ci and C2, and the polar of Pi in Q. Prove that Pi and Q divide the segment C1C2 in- ternally and externally in the same ratio. [Solution : Let Pi(xi, yi) and P2(X2, 2/2) he any two points. Then the point P whose simple ratio with respect to Pi and P2 is X has the co- ordinates ^ _ xi + Xx2 , _ 2/1 + X1/2 If these be substituted in the equation of the ellipse bH^ 4- a^y^ = ajb^ and the resulting equation arranged as a quadratic in X, we have /^4.y22_l^^2 + 2f^2^M_2_l^x + f^' + ^'-lUo. V a2 ^ 62 ) ^ \ a^ ^ b^ J \a^ b^ J The roots of this equation are the simple ratios of the points Pi and P2, respectively, with respect to the points d, C2 in which the line P1P2 meets the ellipse. If the segment Ci, C2 is to be divided internally and ex- ternally in the same ratio the roots Xi, X2 of this equation must be equal numerically, but opposite in sign, i.e. Xi -f- Xo must be zero. The coefficient XIII, § 242] THE PARABOLA 375 of X in the above equation would then be zero, if the theorem to be proved is true. But the condition «2 ^ 62 is precisely the condition that P'z{x2^ y-2) be on the polar, of Pi with respect to the ellipse. The similar proofs for the hyperbola and parabola are left as exercises.] 8. Two points P, Q on the line joining two given points d, C2, are said to divide the segment Gi C-i, harmonically, if they divide the seg- ment internally and externally in the same ratio (i.e. if CxP/PCi = — O1Q/QC2). Show that the result of the last exercise leads to the fol- lowing : The locus of a point Q^ such that a given point P and the point Q divide harmonically the segment joining the points in which the line PQ meets a conic is the polar of P with respect to the conic, or a part of the polar. Which part is it ? (Compare with the similar question in Ex. 5.) Properties of Diameters 9. ProTB that the locus of the mid-points of the chords of a conic drawn parallel to a given chord is a diameter of the conic. [Solution for the Ellipse : Let the equation of the ellipse be b^2c^ + a^y^ = a^b'^, and let the slope of the given chord be m. Then any chord parallel to the given chord isy = mx + k. The abscissas Xi, X2 of the points in which this chord meets the ellipse are the roots of the equation (62 + a2^2)a;2 + 2 a^mkx + a^(k^ - 62) = 0. The sum of the roots of this equation is Xi-}-X2= — 62 -I- a^n^^ The coordinates (x', y') of the mid-point of the chord are then X' = itixi -f X2) = - TT^^, y' = mx' +k= ^'^ 62-|-a2m2' 62 + a2m2 The coordinates x', y' then satisfy the equation y = — (b^x^/(a^m), no matter what the value of k is. The locus of the mid-points of the chords whose slope is m is, therefore, the straight line whose equation is y = — (62x)/(a2m). Since this straight line passes through the center of 376 MATHEMATICAL ANALYSIS [XIII, § 242 the ellipse, the theorem is proved for the ellipse. The similar proofs for the hyperbola and parabola are left -as exercises.] 10. From the result of the last exercise, show how to construct a diameter of any conic, and hence (in case of ellipse and hyperbola) how to find the center, when only the outline of the curve is given. 11. Having given the outline of an ellipse or hyperbola, construct the center. Then show how to construct the principal axes (make use of the fact that the principal axes are axes of symmetry ; a circle drawn with the center of the conic as center and suitable radius will meet the conic in the four vertices of a rectangle whose sides are parallel to the principal axes). Then construct the foci and the directrices. 12. Having given only the outline of a parabola, show how to construct the axis, the focus and the directrix. 13. Show that the tangent drawn to a conic at an extremity of a diameter is parallel to the chords which the diameter bisects. 14. If two diameters of a conic are such that each bisects the chords parallel to the other, the diameters are said to be conjugate; and each is called the conjugate of the other. Prove that if wi, W2 are the slopes of two conjugate diameters of the ellipse h'^x'^ + a'^y'^ = d^h'^, then we have m\m2. = — ly^/a^. 15. Prove that, if wi, mi are the slopes of two conjugate diameters of the hyperbola fe^x^ _ a'iyi — a-h'^, we have WiW2 = h'^/a^. 16. The only conic for which all pairs of conjugate diameters are per- pendicular is the circle. 17. The polars of the points on any diameter of an ellipse or hyperbola are parallel to the conjugate diameter. 18. If one extremity of a diameter of an ellipse Jfix^ -\- a^y'^ = aV)^ has the coordinates (xi, yi), one extremity of the diameter conjugate to the given one will have the coordinates (— yia/b, X\b/a). 19. The area of a parallelogram circumscribed about an ellipse whose sides are parallel to two conjugate diameters is constant and equal to 4 db. 20. Prove that, if a\ and &i are the lengths of two conjugate semi- diameters of an ellipse, a\^ + fti^ = a^ -f h^. 21. Prove that any pair of conjugate diameters of the hyperbola 6%2 _ 052^2 _ 0,252 are also conjugate diameters of h'^x^ — a^y'^ = — aP'h'^. 22. If a diameter of a hyperbola with center meets the hyperbola in P and the conjugate diameter meets the conjugate hyperbola in Q, prove that 0P2 _ 0q^ = (£^ - 62. CHAPTER XIV POLAR COORDINATES 243. Review. Polar coordinates, introduced in §§ 112-114, are often useful in studying geometry analytically. The present chapter is devoted to illustrating some of the principles involved and their applications. EXERCISES 1. What is the locus of points for which p is constant ? 2. What is the locus of points for which 6 is constant ? 3. Show that the points (p, d) and (/s, — 0) are symmetric with respect to the polar axis. 4. Show that the points (/>, 6) and (— p, 6) are symmetric with respect to the pole. 6. Show that the points {p,.d) and (/>, d + 180°) are symmetric with respect to the pole. 6. Find the distance between the points A (2, 45 '^) and B(7, 105°). [Hint. Use the law of cosines.] 7. Prove that the distance between the points (pi, ^i) and (p2, ^2) is Vpi'^ + pi^ - 2 P1P2 cos {62 — di). 244. Locus of an Equation. The locus of an equation in the variables p and is such that : (1) Every point whose coordinates (p, 0) satisfy the equation is on the locus or curve, and (2) A set of coordinates * of every point on the locus or curve satisfies the equation. * Not necessarily every set. Thus, the point (2, 60°) = ( - 2, 240°) is on the locus of p = 1 + 2 cos 6 ; but the second set of coordinates does not satisfy the equation. 377 378 MATHEMATICAL ANALYSIS [XIV, § 244 The curve may be sketched by computing a table of corre- sponding values of p and 6, plotting the corresponding points, and then sketching the curve through them. The amount of work may often be shortened if one makes use of the following obvious rules for symmetry : If a polar equation is left unchanged, (a) ichen 6 is replaced by — 6, the locus is symmetric with re- spect to the polar axis. (b) when p is replaced by — p, the locus is symmetric with re- spect to the pole. (c) ivheyi 6 is replaced by 180° + 6, the locus is symmetric with respect to the j)ole. (d) ivhen 6 is replaced 6?/ 180° — 9, the locus is symmetric ivitJi respect to the line through the pole perpendicular to the polar axis. It should be borne in mind, however, that none of these rules are necessary conditions for sym- metry. Why not ? 245. Illustrative Examples. We shall illustrate the methods of plotting curves in polar coordi- nates by the following examples. Example 1. Discuss and plot the locus of the equation p = 4 cos 6. The locus is symmetric with respect to the polar axis. If we plot points from 0° to 90°, we obtain the upper half of Fig. 215. Then by symmetry we obtain the complete graph given in Fig. 215. Why ? Fig. 215 d 0° 30° 45° 60° 90° p 4 3.5 2.8 2 XIV, § 245] POLAR COORDINATES 379 Example 2. Discuss and plot the locus of the equation p — 4t &m'^ 6. The locus is symmetric with respect to the pole, the polar axis, and the Ime through the pole perpendic- ular to the polar axis. If we plot points in the range ^=0° to 0=90°, and make use of symmetry, we have the complete figure which is given in Fig. 216. e p 0° 30° 1 45° 2 C0° 3 90° 4 Fig. 216 The branches constructed by symmetry should be checked by substituting in the original equation the coordinates of at least one point on each branch. Seri- ous errors may thus be avoided. Example 3. Discuss and plot the locus of the equa- tion p =: 1 -h 2 cos (9. The curve is symmetric with respect to the polar axis. If we plot points from 6 = 0° to 6= 180°, we get the points shown in Fig. 217. Then by symmetry we get the complete graph, or the curve in Fig. 217. Fig. 217 6 0° 30° 45° 60° 90° 120° 135° 150° 180° P 3 14-V3 1 -H\/2 2 1 1-V2 1-V3 -1 380 MATHEMATICAL ANALYSIS [XIV, § 245 EXERCISES Are the following loci symmetric with respect to the pole ? The polar axis ? The line through the pole perpendicular to the polar axis ? 1. p = a cos 0. 5. p- = a cos 2 d. 9. p = a sin2 6. 2. p = a sin d. 6. p'^ = a sin 2 6. 10. P = sin^l p = e. 3. p = a (1 — cos d). 7. p = a cos 2 ^. 11. 4. p = a(l — sini 9). 8. p = a sin 2 ^. 12. p2 cos 0=4. Discuss and plot , thel ocus of each of the following < equations. 13. p = 5. 22. p cos ^ = 4. 31. p =z 1 + 2 sin e. 14. p=- 5. 23. p cos ^ = — 4. 32. p = 1 — 2 cos d. 15. p2 = 25. 24. p sin ^ = 5. 33. P = 1 — 2 sin 0. 16. ^ = 30°. 25. p sin ^ = - 6. 34. p = 2 + cos 0. 17. e=- 30°. 26. p = 1 - cos ^. 35. p = 2 + sin d. 18. p = 8 cos d. 27. p = 1 + cos ^. 36. p = 4 tan d. 19. p = - 8 cos e. 28. p = 1 ~ sin ^. 1 29. p = 1 + sin 0. 37. p — '■ 20. p = 8 sin e. l-cos^* 21. p = — 8 sin e. 30. p = 1 + 3 cos ^. 246. Standard Equations. We shall now derive polar equa- tions for the straight line and the conic sections. The straight line. Let CD be any straight line (Fig. 218) ON = p the perpendicular let fall upon it from the pole 0, and a the angle which this perpendicular makes with the polar axis. Let (p, 0) be any point on the line. ON Then --- = cos (0 — a) or cos (a — 6). But by § 120, cos (0 — a) = cos (a — $). Hence (1) P = p cos (6 - a) is the desired equation. XIV, § 246] POLAR COORDINATES ' 381 If the line is perpendicular to the polar axis, its equation is p cos ^ = p. Why ? If the line is parallel to the polar axis, its equation is p sin ^ = p. Why ? The circle. Let C (c, a) be the center of a circle of radius r and P (pj 6) any point on the curve (Fig. 219). In the triangle Fig. 219 OOP, OC=c, OP = p and the angle COP = ± (0 - a) depending upon the position of the point P. But since cos (0 — a) = cos {a — 0), we have from the law of cosines, §126, (2) r2 = c2 + p2 _ 2 c p cos (6 - a) as the equation of the desired locus. If the center C is upon the polar axis (a — nir), equation (2) becomes (3) r2 = c2 -f p2 ± 2 c p cos d. If the circle passes through the pole (r = ± c), equation (2) becomes (4) p = ± 2rcos(^-a). If the center C is upon the polar axis (a = 0) and the circle passes through the pole (c = ± r), equation (2) becomes (5) p = ± 2 r cos 6. If the center is at the pole (c=0), equation (2) becomes p= ±r. 382 MATHEMATICAL ANALYSIS [XIV, § 246 The Polar Equation of any Conic. The polar equation of any conic may now be derived. Let DD' be the directrix, F the corresponding focus, and e the eccentricity. Let the per- pendicular through F to DD' meet DD' in L. Let the segment LF = p, and take F as the pole and the extension of the line LF as the polar axis. If P(p, 6) is any point on the curve and PS is the perpendicular from P to DD\ then by the definition of a PS, Fig. 220 conic, we have that is, or (6) FP==e p = e{p + p cos 6), 1 — e cos 6 which is the polar equation of a conic. If e < 1, the equation represents an ellipse ; if e = 1, a parabola ; e > 1, a hyperbola. EXERCISES 1. Derive the equation p = 2 r cos [ (5) , § 246] directly from a figure. 2. Derive tiie polar equation of the ellipse assuming the right-hand focus as the pole and the major axis as the polar axis. 3. Derive the polar equation of a hyperbola assuming the right-hand focus as the pole and the transverse axis as the polar axis. 4. Derive the polar equation of the circle which passes through (0, 0°) and has its center at (r, 90°) ; (r, 270^). 6. Derive the polar equation of the parabola assuming the focus at the pole and the directrix the line p sind = p ; psind = — p. 6. The difference of the focal radii of a certain hyperbola is 3, and the distance between the foci is 6. Find a polar equation of the curve. XIV, § 247] POLAR COORDINATES 383 247. Other Curves. What is the advantage of polar coordi- nates ? Why not continue to use only rectangular coordinates ? The answer to these questions is that in certain kinds of problems polar coordinates are much more convenient. The following examples will illustrate the desirability of polar coordinates. The limacon. Through a fixed point upon any given circle of radius a, a chord OPi is drawn and produced to P so that Fig. 221 P^P = kj where A; is a given constant (Fig. 221). Find the locus of P"* as Pi describes the circle. If p — 2 a GosO is the equation of the circle and the pole is the fixed point, then the locus of P is (7) p = 2 a cos 0-\-Jc. If k = 2 a, the equation may be written in the form (8) p = 2 a(l + cos 0) and the curve is known as the cardioid, on account of its heart-shaped form (Fig. 222). fig. 222. - The Lima9on * This curve is known as the lima<^on of Pascal. It was invented by Blaise Pascal (1623-1662) , a famous French mathematician and philosopher. The word lima(;on means snail. The Germans call this curve die PascaVsche Schnecke. 384 MATHEMATICAL ANALYSIS [XIV, § 247 The cissoid. OA is a fixed diameter of a fixed circle (Fig. 223). At the point A a tangent is drawn, while about the point a secant revolves which meets the tangent in B, and the circle in C. Find the locus of a point P on OB so deter- mined that 0P= CB. Fig. 223 Take as the pole and OA as the polar axis of a system of polar coordinates. If we denote OA by 2 a, then the equation of the circle is p= 2 a cos 6. Let P be denoted by (p, 6). Now p=OP=OB-PB. 0B=2aseGe and PB = 0C=2acose. p = 2 a (sec — cos 0), p = 2 a tan ^ sin 0. Fig. 224. — The Cissoid The locus of this equation is given in Fig. 224. The curve is known as the cissoid of Diodes.* * Cissoid (Greek, Kto-o-ds = ivy) means ivy-like. The Greeks considered only the part of the curve lying within the circle. Diocles was a Greek mathematician who lived sometime between 217 b.c. and 70 b.c. By means of this curve, Diocles showed how to construct the side of a cube whose volume is twice the volume of a given cube. See Ex. 4, p. 388. XIV, § 247] POLAR COORDINATES 385 Conchoid of Nicomedes.* A straight line revolves about a fixed point and meets a fixed straight line MN in the point Q. From Q a fixed length is laid off on OQ in both direc- tions. The locus of the two points, P and P', thus determined is called a conchoid. Let be the pole and the line OR through perpendicular to MJSf be the polar axis of a system of polar coordinates Fig. 225. —The Conchoid Let (p, 6) be the coordinates of any position of the generating point P (or P'). Then p = OP(or OP') = OQ±QP=OBseGe± QP. But OR and QP are given constants ; call them a and b respectively. Then (10) p = asec^±6 is the desired equation of the conchoid. * Conchoid (Greek, Koyxo? = mussel) means mussel-like. Nicomedes was a contemporary of Diodes. He invented the conchoid for the purpose of trisect- ing an angle, which is one of the famous problems of antiquity. This prob- lem cannot be solved by means of the compass and straightedge alone. 2c 386 MATHEMATICAL ANALYSIS [XIV, § 248 248. Spiral Curves. A spiral is a curve traced by a point which revolves about a fixed point called the center, but con- tinually recedes from or continually approaches the center ac- cording to some definite law. The spiral of Archimedes is the locus of a point such that its radius vector is proportional to its vectorial angle. Therefore its equation is (11) p = ke, where Ic is a constant.* The form of the equation shows that the locus passes through the pole, and that the radius vector increases without limit as the number of revolutions increases without limit. Figure 226 represents a portion of the locus for k = ^j with $ expressed in degrees. The hyberbolic or reciprocal spiral is the locus of a point such that its radius vector is inversely proportional to its vec- * In this example, and in those that follow, it is usual to express the angle in radians ; but this is not necessary, since the same result can be obtained by choosing a different value for A: if d is exi>ressed in degrees. XIV, § 248] POLAR COORDINATES 387 torial angle. The equation of the locus is therefore (12) P=^. where A; is a constant. Figure 227 represents a portion of the graph for A; == 70 and for positive values of 6 (expressed in degrees). The logarithmic spiral is the locus of a point ^^^- ^27 such that the logarithm of its radius vector is proportional to its vectorial angle, i.e. (13) log p = kO, where A; is a constant. If the base of the system of logarithms is b, the equation may be written in the form p = 6*^. Figure 228 represents a portion of this locus when 6 = 3, for A: = y^, with 6 expressed in degrees. 388 MATHEMATICAL ANALYSIS [XIV, § 248 EXERCISES 1. Discuss the form of the limagon (7), when | A: | < 2 a. When |A:|>2a. 2. Solve the locus problem used to define the lima9on by means of rec- tangular coordinates, and compare the merits of the two solutions. 3. By taking the line OA (Fig. 223) as the ic-axis and the tangent to the circle at O as the y-axis, prove that the equation of the cissoid is 2a-x 4. Duplication of a Cube. In the adjoining figure, let MN= a and MB = 2 a. Draw BA and let it meet the cissoid in the point D whose ordinate is LD. Prove that ZZ)3 = 2 OLK If MB = n • a prove that LD^ = n ■ OL^. 6. If in Fig. 225 the line MN^ is taken as the a;-axis and the line OB A as the jz-axis, prove that the equation of the conchoid is a:V=(y + «)'(&' -2/2). Compare the merits of this solution with that on p. 385. 6. Trisection of an angle. Let AOB be the angle to be trisected. Through a convenient point A on one side OA of the angle draw AB per- pendicular to OA. Through B draw a line B C parallel to OA. From as fixed point, and AB as fixed line, and 2 . OB as a constant dis- tance, describe a conchoid meeting BC in C. Angle ^OC is then \ AOB. [Hint: E is the mid-point of Z>C ; then OB = BE = EC. The result then follows from elementary geometry.] 7. Show that in the conchoid, if (a) 6 > a, the curve has an oval at the left, as in Fig. 226; (6) 6 = a, the oval closes up to a point ; (c) 6 < a, there is no oval and both branches lie to the right of the point 0. XIV, § 249] POLAR COORDINATES 389 8. Draw the parabolic spiral, which is defined by the equation p^ = kd. Take k = -^^ with 6 in degrees, and use only the positive values of p. 9. Draw the lituus, which is defined by the equation p^ = k/d. Take A; = 90 with 6 in degrees, and use only the positive values of p. 249. Relation between Rectangular and Polar Coor- dinates. Take the origin of a system of rectangular axes as the pole, and the positive half of the a;-axis as the polar axis, of a system of polar coordinates. Let (x, y) and (p, 6) be respectively the rectangidar and polar coordinates of any point P. Then x/p — cos ^, yjp — sin 6. Hence, we have I X = p cos e, I y = p sin e. (14) It is here assumed that the coordinates of P are so chosen that OP — p and angle XOP = 0. This is always possible. If p is positive, X always has the sign of cos and y the sign of sin $, ft<=^' -?^ F P Fig. 229 Conversely, if p is positive, we see from Fig. 229 that .2 _ ^2 _L „2 gin $ — ^ (16) x" + y^, $ = arc tan [t\ cos d = ■'- ' 390 MATHEMATICAL ANALYSIS [XIV, § 249 EXERCISES Transform the following equations into equations in rectangular coordi- nates. State in each case whether the graph is easier to sketch from the polar or the rectangular equation. 1. p = 1 - cos 0. 6. p2 sin 2 e = 3. 11. p^ = 6. 2. p=l + sin^. 7. p2cos2^ = 4. ^^ ^2^1 3. p = 4. 8. p2 = cos 2 e. 13. p = a sec d-\- b. 4. pcos^ = 5. I 14. p = 2asec^tan^. 10. p = — • 6. p sin ^ = - 2. e 15. p = 4 cos 2 6. Transform the following equations into equations in polar coordinates : 16. ^2 + 1/2 = 4 X. 21. xy = 4. 17. (x2 + y2)2 _aj2 _ 2/2. 22. a; cos a 4- y sin a = ;>. 18. X -y = 0. 23. (y2 4- x2 _ 2 a;)2 = x2 -f- y2. 19. 1/2 = 4 X. 24. x3 = ?/2(2_a;). 20. 9 x2 + 4 ?/2 = 36. 25. x2?/2 = (y + 2)2(4 - y^). MISCELLANEOUS EXERCISES Sketch the follow fving curves : 1. p = a cos 2 d. 1. p = a cos 6 6. 2. p = a cos 8 d. 8. p = a sin 5 ^. 3. p = a sin 2 ^. 4. p = a sin 3 6. • 6 9 p = a sin-- h. p = a cos 4 ^. 6. p = a sin 4 ^. 10. p = acos-- Find the points of intersection of the following pairs of curves. Plot the curves in each case and mark with their respective coordinates the points of intersection. 11. p == 1 4- cos e,. 14. p = 1 + cos d, 4(1 +cos0)p = 1. 2p = 3. 12. p = 4, 15. p = 2(l -sin^), pcos^=2. (l+sin^)p=l. 13. p = \/2, ^®- P = cosd, P = 2 sin 5. P = 1+ 2 cos d. XIV, § 249] POLAR COORDINATES 391 Solve the following exercises by the use of polar coordinates : 17. Find the locus of the center of a circle which passes through a fixed point O and has a radius 2. 18. Prove that if from any point a secant is drawn cutting a circle in the points P and Q, then OP - OQ is constant for all positions of the secant. [Hint. By using equation (2), §246, show that the product of the roots is constant.] 19. Secants are drawn to a circle through a fixed point on the cir- cumference. Find the locus of the middle points of their chord segments. 20. Find the locus of the middle points of the focal radii issuing from one focus of an ellipse ; parabola ; hyperbola. 21. The focal radii of a parabola are produced a constant length. Find the locus of their end-points. 22. Through a fixed point on a fixed circle a variable secant OP is drawn cutting the circle in B. If BP = 3 OB, find the locus of P. CHAPTER XV PARAMETRIC EQUATIONS 250. Parametric Equations. As a point P(xj y) moves along a given curve, the x- and y- coordinates of the point vary. So do many other quantities connected with this point, as for ex- ample, in general, its distance OP from the origin, the angle which OP makes with the aj-axis, its distance from a fixed line, etc. It is sometimes convenient to express x and y in terms of one of these variables. This third variable, in terms of which the variables x and y are expressed, is called a parameter. For exaonple, we see that the coordinates of any point P{x, y) on the circle whose center is at the origin and whose radius is r, can be expressed in the form (1) rx = rco I 1/ = r si r cos 0, sin 6, Fig. 230 where 6 is the angle XOP (Fig. 230). These are then para- metric equations of the circle. If we eliminate the parameter 6 between these two equations by squaring and adding them, we obtain the equation 2 _j_ ,,2 _ ^2 X -i- y — r , which is the rectangular equation of the circle. 392 XV, § 250] PARAMETRIC EQUATIONS Similarly, a pair of parametric equations of the ellipse (2) are (3) 393 ^4-^ = 1 a2 62 f X = a cos 6, 1 y = 6 sin 6 ; for, these values of x and y are seen to satisfy equation (2) for all values of 6. The geometric interpretation of equations (3) is important. In Fig. 231, a geometric construction given in § 226 is used. Fig. 231 The abscissa of P is equal to the abscissa of Q, i.e. a cos 0, the ordinate of P is equal to the ordinate of R, i.e. b sin 6. There- fore the coordinates of P are x = a cos 0, y = b sin 0. The angle is known as the eccentric angle of the point P. We should note that is not the angle XOP. A pair of parametric equations for the hyperbola (4) ^^-yl=i ^ ^ a^ b^ are (5) x = a sec 9, y = b tan 9, for these values of x and y satisfy the equation (4). It is important to note that a given curve may have as many sets of parametric equations as we please. For example, para- 394 MATHEMATICAL ANALYSIS [XV, § 250 metric equations of a circle may be written in the form x = a cos t, y = a sin t, as above ; or they may be written in the form or in any one of many other forms. EXERCISES 1. Show that x = t, y =2 — t are parametric equations of a straight line. 2. Show that x = I pt^, y =pt are a pair of parametric equations of the parabola y'^ = 2 px. 3. Write two pairs of parametric equations for the line y = x. 4. Prove that ., ,„. „ , are parametric equations of a circle. 6. Write a pair of parametric equations for the rectangular hyperbola a;2 - ?/2 _ ^2, 6. Show that x = Acosd -\- Bsin d, y = Asm 6 — Bcosd are para- metric equations of a circle. 7. Prove that x = 6 + 4 cos ^, y = — 2 + 3 sin ^ are parametric equa- tions of an ellipse. 8. Write a pair of parametric equations for the circle (x - a)2 + (y - 6)2 = r2. 9. Prove that a; = 6 + 4sec^, y=— 2 + 3 tan d are parametric equa- tions of a hyperbola. 10. Find the equation of the tangent to X^ f/2 (<^) -^ + 7^ = ^» *^ ^1 = « <^08 ^i» yi = ft sin ^1- /v2 «f2 (ft) — o — t:; = 1' at ^1 = « sec 01, yi = 6 tan ^i. (c) ?/2 = 2px, at xi = i ;)«i2, yi = pti. XV, § 251] PARAMETRIC EQUATIONS 395 11. Prove that the tangents to y''- = 2px at {^ph^.ph)^ ilPh^.Ph) meet at the point [^^phti, \p{h + ^2)]- 12. Write the equation of the tangent to y^= 4 ax at xi = 13. Show that Wli^' 2a_ 3a« y = 3a«2 1 + «3 ' ^ 1+^3 are parametric equations of the curve x^ -{-y^ —Z axy. 14. Show that x = a cos^ e, y = a sin^ d are parametric equations of 21 2 the curve x'^ -i- y^ = a^. 15. Find the x- and ^/-equations of the curve whose parametric equa- tions are X = a (^ — sin d), y = a (1 — cos d). 251. Sketching Loci of Parametric Equations. If we assign a series of values to the parameter and determine the series of corresponding pairs of values for x and 2/j we can interpret these values as the coordinates of points on a curve. Plotting these points and sketching a curve through them, we have the graph of the curve whose parametric equations were given. Example. A pair of parametric equations giving the path of a body projected horizontally from a height of 400 ft. with a velocity of 10 ft./sec., are x=10t, y = 400 — 16 t^. Sketch the locus. t 1 2 3 4 5 X 10 20 30 40 50 y 400 884 336 256 144 Y 400 300 800 100 N \ \ \ V \ 1 a s 4 6 6 X Fig. 232 In the preceding table are given the values of x and y corresponding to the integral values of t from to 5 inclusive. Plotting these points we have the graph in Fig. 232. This curve is of course the same that we should obtain by first elimi- nating t and then plotting from the equation in x and y. 396 MATHEMATICAL ANALYSIS [XV, § 252 252. The Time as Parameter. Suppose a point moves in a plane. At every instant of time t the point occupies a certain position (x, y). In other words, the coordinates x and y of the point P are functions of t, i.e. ,/,x r a; = a function of ty I y = a function of t. These equations are then parametric equations of the path traversed by the point. Such equations arise frequently in mechanics when it is desired to describe the motion of a body subject to various forces. For example, if a body is projected from a point (0, 0) in a vertical plane at time i = 0, with an initial velocity Vq, and making an angle a with the horizontal (ic-axis), its position at the end of t seconds * is given by the equations {x = tv^ cos a, [y = tVoSma-}gt% where, if Vq is measured in ft./sec, gr is a constant approxi- mately equal to 32.2. The use of these equations of a projectile will be illustrated in the next article and the exercises follow- ing it. EXERCISES Sketch the following curves from their parametric equations. 1. x = t, 4. X = t'^ - 1, 7. x = t, lo. x = t, 2/ = « + 2. 2. x = r^, y = t^-\. 5. X = ^2 4. 1, 1 11. y = t-tK X = sin d, y = r. y = z. 8. x = t\ y = cos d. 3. x = s+ 1, 6. x = t, y = t\ 12. x = tan e, y = s'^. y-\- 9. x = t^+\, y = t^^l. y = sec d. 13. x = \Qt cos SO'', 14. x = 5«, ?/ = 25 + < sin 30° - 16 «2. y = so- 16 «2. * The resistance of the air being neglected. XV, § 253] PARAMETRIC EQUATIONS 397 253. The Path of a Projectile in Vacuo. The equations (7) given in § 252 yield many results of interest regarding the paths of pro- jectiles. Some of these are given in this article and others are found in the exercises below. They are, of course, only approximations to the actual behavior of a projectile, in view of the fact that the resistance of the air has been neglected * By eliminating t between the equations (7), § 252, we obtain the equa- tion of the path in rectangular coordinates (x, y) : (8) y = X tan a — gx^ Fig. 233 The path is, therefore, a parabola, with a vertical axis. The vertex of the parabola is at the point (Fig. 233) (9) \m 2 a, <''''''' ' ^9 The greatest height above the horizontal is (10) H = The complete range, i.e. the distance from O to the point where the projectile again meets the horizontal, is found as follows : If in (7) we place y = 0, we find t = and t = 2{vjfj) sin a. The value of x for the second value of t found is the desired range B, i.e. .2 B ^ sin 2 a. g This result could also have been found by placing y = in (8) and solv- ing for X. Why ? * For the theoretical and practical discussion of the flight of actual projec- tiles (whose motion is 'appreciably affected by the resistance of the air) the student is referred to Alger, The Groundwork of Naval Gunnery, or External Ballistics. 398 MATHEMATICAL ANALYSIS [XV, § 253 EXERCISES 1. A gun is fired at an elevation of SO'^. Find the range if the muzzle velocity of the shell is 1000 ft./sec. Aiis. About 5 mi. 2. What is the greatest height reached by the projectile in Ex. 1 ? How long is its time of flight ? 3. What must be the initial velocity of a baseball thrown at an angle of 45° in order that it may travel 200 ft. before hitting the ground ? 4. A stone is thrown from a tower 100 ft. high, with an angular ele- vation of 45° and an initial velocity of 64 ft./sec. How far from the foot of the tower will the stone hit the ground ? 6. The great pyramid of Cheops is 450 ft. high. Its base is a square 746 ft. on a side. A ball is thrown upwards from the top in a direction making an angle of 20° with the horizontal and with the velocity of 80 ft./sec. Will the ball clear the base of the pyramid ? 6. Prove that for a given initial angle of elevation the range of a pro- jectile is proportional to the square of the initial velocity. 7. Prove that for a given initial velocity the maximum range is obtained when the angle of elevation is 45°. 8. Prove that with the notation of § 253, the time of flight of a pro- jectile from to (x, y) is (x/vo) sec a ; from Oto (J?, 0) is (2 Vo/g) sin a. 9. Prove that the paths of a projectile with given vq, but varying a, have the same directrix. 10. Prove that the coordinates of the focus of the path of a projectile are ( vq^ sin 2 a — ih)^ cos 2 a \ 2sr ' 2g )' Hence show that the locus of the foci of all paths in a given vertical plane with the same vo is a circle with center at 0. 11. Prove that the parabola of maximum range has its focus on the avaxis. 12. Prove that the locus of the vertices of the paths with given "yo is an arc of an ellipse. 13. Prove that the locus of the vertices of the paths with a given a. and a varying vo is a straight line. 14. Prove that the locus of the foci of the paths with a given a and a varying vo is a straight line. XV, § 254j PARAMETRIC EQUATIONS 399 H ^ 254. Locus Problems. Parametric equations of a curve are sometimes much more easily obtained and easier to work with than either the rectangular or polar equations. The following problems illustrate some of the methods that may be em- ployed. Example 1. A line of fixed length moves so that its ends always remain on tlie coordinate axes. Find the locus gen- erated by any point of the line. Call the point whose locus is desired P(x, y). Since the line is of fixed length, call the segments into which P divides it, x = a cosd^ y=b sin d. Example 2. a fixed line. Take for origin the point O where the moving point P touches the fixed line. If r is the radius of the circle and the angle PCD (Fig. 235) is 6 radians, then PD = rsind, DC = r cos 6 and OB = arc BP = rd. Fig. 234 a and b (Fig. 234). Then Therefore the locus of P is an ellipse (§ 250). Find the locus of a point P on a circle which rolls along Fig. 235 Now if P is denoted by the coordinates (aj, y), x= 0A= OB- AB= OB- PD = rd- r sine = r(d- sin d), y = APz=BC- DC = r-r cos d = r{l- cos d). Therefore ni\ {x = r(6— sin^), \y = r(l — cos d) are parametric equations of the curve traced by the point P. This curve is known as the cycloid. 400 MATHEMATICAL ANALYSIS [XV, § 254 Example 3. Find the locus of a point P on a circle of radius a which rolls on the inside of a circle of radius 4 a. Take the center of the fixed circle as the origin and let the x-axis pass through a point M where the moving point P touches the large circle. Let angle MOB = d radians. Now we have arc PB = arc MB = 4 ad and arc PB = a x angle PCB. Therefore a X angle PCB = 4 ad, or angle PCB = 4 ^. But Z OCD + ZDCP-^Z PCB = T. Therefore |-0 + Z2)CP+4d = T, i.e. ZDCP=--Se. Fig. 236 Now if the point P is denoted by {x, y) we have x = OE=OD-\- DE = OD-\- FP = OCcose + CP&ml--Z e\ = 3 a cos ^ + a cos 3 ^ = 4 a cos* ^,* y = EP=DG-'FC= OC sine - CPcosl--Se\ = 3 a sin ^ — a sin 3 ^ = 4 a sin* ^ ;♦ that is, (12) a: = 4acos3^, y = 4a8in^0. This curve is called the four-cusped hypocycloid. Example 4. P8P' is a double ordinate of an ellipse ; Q is any point on the curve. If QP, QP' meet the x-axis in O and 0', respectively, prove that CO • CO' = a^, where C is the center of the ellipse. Let P be (a cos ^i, ft sin di), then P' is (acos^i, — ftsin^i). Let Q be (a cos 6, b sin d). The equation of line PQ is y " sin a = —^ -(X — a cos a) a(cos^i — cos^) * Prove that cos 3 ^ = cos (2 + ^) = 4 coss — 3 cos ^, sin 30 = sin (2 + ^) = 3sin0— 4 8in«tf. XV, § 254] PARAMETRIC EQUATIONS 401 and its x intercept {i.e. CO) is a(cos 6 sin d^ — sin 6 cos 6i) sin d — sin di Similarly CO' is «( - cos ^ sin ^i - sin cos gQ , sin ^ + sin di The product CO • CO' gives a"^. EXERCISES 1. A line of fixed length 2 a moves with its ends always remaining on the coordinate axes. Find the locus of the mid-point of the line. 2. Find the locus of the middle points of chords of an ellipse drawn through the positive end of the minor axis. 3. Find the locus of a point P' on the radius CP of the cycloid (Fig. 235) if CP' = b and b<r. 4. The same as Ex. 3, except & > r. 6. A circle of radius r rolls on the inside of a circle of radius a. Find the locus of a point P on the moving circle. a — r Ans. The hypocycloid X = (a — r) cos d -{- r cos r a — r = (a — r) sin ^ — r sin — - — < Ans. The epicycloid X = (a -\- r) cos 6 — r cos y = {a -h r) sin ^ — r sin where 6 is the same as in Example 3, § 254. 6. A circle of radius r rolls on the outside of a circle of radius a. Find the locus of a point P on the moving circle. a + r a + r —re, where 6 is the same as in in Example 3, § 254. 7. The area of the triangle inscribed in an ellipse, if ^i, ^2* ^3 are the eccentric angles of the vertices, is \ ah [sin (^2 — ^3) + sin (^3 — ^1) + sin (^1 — ^2)] ^-2 ah sin ^2ii_^ sin h=Jl sin ^J-ZLb, 2 2 2 8. The coordinates of one extremity of a diameter of an ellipse are (a cos ^1, ft sin ^1). Show that the coordinates of one extremity of the conjugate diameter are (— a sin ^1, h cos ^1). 2d PART IV. GENERAL ALGEBRAIC METHODS THE GENERAL POLYNOMIAL FUNCTION CHAPTER XVI MISCELLANEOUS ALGEBRAIC METHODS 255. The Need of other Methods. We have hitherto con- sidered special functions such as a;2, sin ic, log^QX, or special types of functions such as mx -{- h, ax^ + bx -{-c, log„ x, a'' ; and we have studied their geometric and other applications. The study of more general types of functions, for example, the general polynomial of degree n, a^x"" + a^_iX'^~^ -h h «i^ + «o) of which mx + h, ax^- -f- 6a3 -f- c, ay? + bx^ -\- cx^- d are special types, requires more powerful methods. Some of these we propose to consider in the present and the succeeding chapters* 256. Technique of Polynomials. We shall first recall the technique of the addition and multiplication of polynomials. We begin by noting that a polynomial can be completely represented by so called detached coeffi- cients, as follows : « ^ ^ « the place of each coefficient in this expression indicating the power of X to which it belongs. Thus, for example, 2-316 402 XVI, § 256] ALGEBRAIC METHODS 403 represents the polynomial 2oi:^ — Sx^ + x-\-6; and 10 0-1 represents the polynominal x^ — 1.* To add two or more polynomials we need merely add the coefficients of like powers of x. Thus, the sum of x"^ — l,x*-\- 2s(r^-\-4:X^ + Sx-\-5, and 2a^ — 5x'^-\-x-\-lis given by 10-1 12 4 3 5 2-5 1 1 1 4 4 5=a;* + 4a^-f4a; + 5. The analogy of this process with that of adding a column of numbers may be noted. The product of two polynomials A and B is obtained by multiplying A by each term of B and adding the results. Why ? The multiplication of two polynomials by the method of detached coefficients is also quite analogous to the familiar method of multiplying two integers. Thus the product of 2 a^ -\- Sx"^ — x —2 by aj2 + a; + 4 is given by 23 -1 -2x114 2 3-1-2 2 3-1-2 8 12-4-8 2 5 10 9-6 -8=2a^-\-5x'-\-10a^-\-9x^-6x-S. The student should convince himself, by multiplying these polynomials in the ordinary way, that the above method is indeed valid. ♦This method of representing polynomials will seem very natural, if we note the analogy with the familiar method of representing integers. The number 217 is simply a short way of writing 2 x 102 + 1 x 10 + 7, i.e. the value of the polynomial 2z^-{-x + 7, when x = 10. We have, therefore, been famil- iar with the method of detached coefficients from the time when we first learned to write numbers. 404 MATHEMATICAL ANALYSIS [XYI, § 257 257. The Division Transformation. A polynomial B is said to be a factor of a polynomial A, if there exists a poly- nomial Q such that A = BQ. A is then said to be divisible by B. If no such polynomial Q exists, and if the degree of B is less than that of A, we may always determine polynomials Q and E, such that (1) A = BQ-\-E. Furthermore, the remainder R can always be so determined that its degree is less than the degree of B. The process whereby a given polynominal A is expressed in terms of another polynomial B in the form (1), i.e. the process of finding Q and R, when A and B are given, is called the division transformation. That it is always pos- sible to find polynomials Q and R, if the degree of B is less than that of A, will be clear from the consideration of the following example. Example. Given A = 2x* + 5x^— 'Jx'^ + 12x — 5 and B = x^-10x + 8 to find a polynomial Q such that ^ — J5^ is a polynomial of degree less than that of B. Since the term of highest degree in J. is 2 x* and that in B is x^, it ap- pears that A — 2 x^B can contain no term of degree higher than 3. In fact, we find A — 2 x^B = 25 x^ — 23 x^ -f 12 x — 5. Similarly, since the term of highest degree in ^ — 2 x^B is 25 x^, we see that the expression {A — 2x^B)— 25 xB can contain no term of degree higher than 2. By continuing this process we shall arrive at a polynomial which is of degree less than that of B. The work may be arranged as follows. ^=2x* + 6 x8- 7x2-1- 12x J5. 2x2 = 2x4 -20x3 + 16x2 x2-10x + 8 = 5 2 x2 + 25 X 4- 227 ^-5.2x2 = J5-25x = 25x8- 23x2-}- 12X-6 25x8-250x2+ 200 X ^-^(2x2 + 25x) = J? . 227 = ?(2x2 4-25x + 227) = 227x2- 188X-5 227x2-2270x+1816 2082x-1821 = 7 XVI, § 258] ALGEBRAIC METHODS 405 If the meaning of each step in the process is followed by means of the expressions written at the left,* it will be seen that the process has deter- mined a polynomial ^ = 2^^ + 23x + 227 such that A— BQ = B, where B is of degree less than that of B. The nature of the process shows that finally such a polynomial B will always be reached. 258. Remarks on the Division Transformation. While the process discussed in the last article is known as the division transformatioyi, it is not a process of division. Only if we take a further step and divide both members of the identity (1) (§ 257) by B, to obtain (2) |=0 + |, do we really divide A by B. The importance of this distinction lies in the fact that the relation (1) as derived above is valid without distinction for all values of the variable x involved. For in deriving the relation we made use only of the opera- tions of multiplication and subtraction. However, the relation (2) becomes meaningless for all values of x for which 5 = 0. We assumed in deriving the relation (1) that the degree of B was less than that of A. This is indeed necessary if Q is to be a proper polynomial. However, if the degree of B is equal to that of A, the same process will lead to a relation (3) A = B'q-\-R, where g is a constant. If the degree of B is greater than that of A, we may obviously write the trivial relation (4) A=B'()^A where A equals R and is by hypothesis of lower degree than B. If we consider a constant as a poljoiomial of degree 0, the last * These expressions are not of course a necessary part of the process. They are given here only to facilitate understanding. 406 ' MATHEMATICAL ANALYSIS [XVI, § 258 two cases may be included in the form (1). Our theorem then takes the general form Given any two polynomials A and B of degrees greater than 0, the7i polynomials Q and R can always he found such that for all values of the variable we have A = BQ -{- R where R is either zero or of degree less than that of B. Moreover, the transformation of A to the form BQ-\- R is unique, i.e. there exist just one polynomial Q and one poly- nomial R satisfying the conditions of the theorem. For, suppose there were a second pair, for example Q' and R'. We should then have BQ+ R = B'Q' + R', or B{Q - Q') = R' — R. But R' — R is either zero or of degree less than that of B, while B(Q— Q') is either zero or degree equal to or greater than that of B. Hence both are equal to zero and R = R', Q=Q'. EXERCISES 1. Add the following polynomials by means of detached coefficients : (a) 2ic2 + 7x + 1, 5a;2 + 2, 3x3 + 4x-8. (6) 6 «2 + 5 ( + 1, 9 <2 + 8 « + 3, 6 «3 + 2 « + 1. (c) ay^ -\- by + c, 2 ay^ + 3by + i, 3 ay^ + 6by + 7 c. (d) 4 a2 -f 3 a + 2, 6 aM- 1, 4 a2 + 2 a + 3, 2 a2 + 6 a. 2. Perform the following multiplications by means of detached coeffi- cients : (a) «3 + 2x2 + X + 3 by 2 X + 1. (c) a* + 1 by a'^ + 1. (b) x8 + 3x2 + 4 by x3 + X + 2. {d) y^ + 1 y + \2\)y y^ + 3y + 2. 3. In each of the cases below transform A into the form BQ ->r B, where B is of degree less than B. Also write down the corresponding form for A/B. Detached coefficients may be used to advantage. (a) ^ = 6x4 4- 7 x3 - 3 x2 - 24 X - 20, B = 3 x2 + X - 6. (6) ^=3x4 + 2x3-32x2-66x-35, £ = x2-2x-7. (c) J[ = 2 x6 + 5 x8 + 13 x2 + 2 X, 5 = x2 -f 2 X + 4. (d) ^ = 4x7-^3x^- 19x* + 2x3-4x2-4x4-7, B = x8-x-6. XVI, § 259] ALGEBRAIC METHODS 407 4. Determine m and n so that k* — moc^ -^-x"^ - nx + \ may be exactly- divisible by a;2 + 2 X + 1- 6. Prove the following propositions : (a) If we multiply the dividend A by any constant as k{k ^ 0), we multiply the quotient and the remainder by k. (6) If we multiply the divisor by A; (A; ^t 0), we divide the quotient by k but leave the remainder unchanged. (c) If we multiply both dividend and divisor by k{k ^ 0), we multiply the remainder by k but leave the quotient unchanged. 259. The Highest Common Factor of two Polynomials. Two polynomials A and B may or may not have a common factor of degree greater than 0, i.e. a polynomial F (of degree greater than 0) may or may not exist such that A = FQ^ B = FQ' where Q and Q' are also polynomials. If no such polynomial F of degree greater than exists, then A and B are said to be prime to each other. If, on the other hand, they have a common factor of degree greater than 0, the one of the highest degree is called the highest common factor (H. C. F.). Theorem 1. If A and B are polynomials with a common factor and M and N are any two polyyiomials, then any common factor of A and B is a factor of AM-\- BN. For let F be any common factor of A and B. Then we have A=::FQ and B=FQ'. Therefore AM-\- BN= F(QM+ Q'N), which shows that i^ is a factor of AM + BN. Theorem 2. If A, B, Q, R are polyyiomials such that A = BQ-\- B, the common factors of A and B are the same as the common factors of B and R. For, by Theorem 1, any common factor of B and i2 is a factor of A and hence a common factor of A and B. More- over, from the relation A — BQ = R and the last theorem, any common factor of A and B is a fac- or of B and R. 408 MATHEMATICAL ANALYSIS [XVL § 259 Successive applications of the division transformation there- fore enable us to find the H. C. F. of two polynomials A and B as follows : Using the division transformation on A and B, we may ™*^ A = BQ + B, where the degree of R is less than that of B. li B = 0, B is the H. C. F. If 72 is a constant different from zero, then A and B have no H. C. F. Why ? If the degree of B is at least equal to 1, we may use the division transformation on B and B to obtain B = BQ,-hB„ where Bi is of degree less than B. If Bi = then B is the H. C. F. of ^ and B. If Bi is a constant different from zero, A and B are prime to each other. If B^ is of degree at least equal to 1, proceed as before, expressing B in the form B = B1Q2 -{- B2. This process may be continued until a remainder B^ is reached which is either zero, or a constant different from zero. If B^ = 0, then B,^_i is the H. C. F. sought. If B^ is a constant different from zero, then A and B are prime to each other. Example 1. Find the H.C. F. of 4a:8_3a;2+7a;-l and 2 a;2-3a;-|-l. A = ^x^ -Sx'^ + I x-1 4a;8-6x2 + 2 x 2x^-3x + l = B 2x + ^ = Q 3 x2 + 6 x-1 3x2 -f a; + | Replace Bhj x — ^% — E'. B = 2x^-3 x + l \x-j^-g=B 2x:2-\%x \ 2x-\i -nx + VA Therefore A and B are prime to each other. XVI, § 260] ALGEBRAIC METHODS 409 Example 2. Find the H. C. F. of x^ - 2 x2 -f x + 4 and 3 x^ -f 8 x2 + 3x-2. The work may be arranged as follows. Since the H. C. F. of two polynomials is not altered in any essential way by multiplying or dividing either of them by any constant (^^ 0), we can avoid fractional coefficients. X3 X3 -2x2 + x-f4 X -2x2 -2x2 + 2x + 4 4-2 21 2x + 2 X4-1 3x3 4- 8x2 4- 3 X- 2 3x3 — 6x^ + 3 x + 12 14x2 -14 14 X2 -1 X2 + x — X -1 — X -1 Therefore the H. C. F. is x 4- 1- EXERCISES Find the H. C. F. of each of the following pairs of expressions : (a) (&) (c) (d) (/) x3 4-2x2-13x+ 10, x3 + x2- 10x4-8. 3x* 4- 5x2 + 2, x6 - 4 X* 4- 5 x2 - 2. a;8_2x2-22x + 8, x2 -6x4- 2. a3 + 3 a2 - 3 a - 5, aa a' 4- 3 a 4- 5. y^ — yf^ — y + i? y'^ -\- y + i- &4 + 5.3 ^ 6 ^-2 _^ 5 5 ^. 5^ ^5 4. 4 53 _^ 52 _ 1 4- x-x2- 5x3 + 4xS 1-4x84-3x4. 6 6 4-1. {h) 4x4-5x3 + x + 1,3x4-4x3+1. (i) a^ 4- a3 + a 4- 1, a2 + a + 1. (j) x^>-l,x-l. 2. Prove that, if the coefficients of two polynomials are rational (or real), the coefficients of the H. C. F. are rational (or real). 3. If F is a factor of A but not of B, how does the H. C. F. of ^4 and FB compare with the H. C. F. of A and ^ ? In introducing and suppress- ing factors during the process of division, what precaution must be taken and why ? 260. Functional Notation. We have already used special notations to represent special functions. Thus sin, cos, tan, log, etc. are special notations with which we are familiar. We shall now introduce a notation that is more general, for it is 410 MATHEMATICAL ANALYSIS [XVI, § 260 applicable to all kinds of functions. We shall use the symbols f{x)j F(x), (t>{x), — to represent various functions of x, and can then speak of the "/-function," " (^-function," etc., just as we speak of the sine function, logarithmic function, etc. Moreover, such a notation can be used to represent the value of the function in question for a given value of the variable. Thus, if f(x) is used to represent the function x^ -{- 2x — 1, the symbol /(2) denotes the value of this function when x = 2 (just as sin (7r/2) means the value of sin x, when x = 7r/2) ; i.e. /(2) =22 + 2. 2-1 = 7. Similarly, with the meaning just given tof(x), we should have f(x -}-h) = {x + hy -^2ix + h)- 1. It should be noted that, when a certain function is called f(x), then, throughout any discussion where this function is used, f(x) always means that particular function and no other. EXERCISES 1. Given/(x) = 3x2 + 2a;- 4, fiiid/(2),/(^),/(0),/(a; + 1/x). 2. Given 0(x) = x/(x - 1), find 0(2), (f>(x + h), 0(1 - x), 0(10). 3. Given F{x) = e* + e-^ find F(0), F(l). 4. Given/(x) = (x - l)/(a; + 1), prove that 6. If 0(x + y) = (t>{x) + 0(j/), show that 0(3 x) = 3 0(a;). 6. Given f(x) = 2 x Vl — x^, prove that /(sinx) =/(cosx) =sin2x. 1 X Given /(x) = -, find the value of /Ml+^V X 8. Given ^(x) = e* + e"*, prove that e{x^y)d{x-y) =^(2x) -}-^(2y). 9. Express tlie fact that the volume of a sphere is a function of its radius. XVI, § 262] ALGEBRAIC METHODS 411 10. Given F {x) = (x- \/x){x'^ - \/x'^){xJ^ - \/x?), prove that F{z) = -F{\/z), 11. Given that/(w) = n !,* prove that (n + 1) f{n) =f(n + 1). 12. Given that/(.7;)=a;2-f-2, find/[/(x)]. 13. Given /(x) = sinx, find/(7r/2),/(7r/3), /(tt). 14. Given /(x) = sin x and <f)(x) = cos x, prove that («) lf(x)T + [0(^)]-^ - 1. (d) fix) = «/,(7r/2 - X). (b) f(x) - 0(a-) = tan X. (e) /(- x) = -/(x). (c) /(.r + 2/) =f(x) cf>(y) +f(y) <p{x). (/) 0(x) = 0( - x). 16. If /(x) = loga X, show that (a) / (a;) + / (y) =f{xy). (c) /(xA) + f(y/x) = 0. (6) /(x") = n/(x). 16. What functions may/(x) represent when (a) fix + y) =f(ix)f(y). (c) /(x") = n/(x). (6) /(x+ 2/) =/(x) +/(2/). (d) /(^^ =/(2/) -/(x). 261. The Remainder Theorem. If a polynomial f(x) is divided by x— a, the remainder isf(a). If f(x) is the dividend, a; — a is the divisor, Q(x) the quotient, and R the remainder, then (5) f{x)=(x-a)Q(x)-^E where B, since it is of lower degree than x — a, does not in- volve X at all, i. e. E has the same value for all values of x. Since the values of the two members of this identity are equal to each other for all values of x, we have for the par- ticular value X = a. ^/ s r> J {a) = M. 262. Factor Theorem. If f(x) is a polynomial and a is a number such that f (a) = 0, then x — a is a factor off(x). The proof of this theorem is left as an exercise. *n! = l-2.3.4 •... • w, that is, 2! =2, 3! =6, 4! =24, .... 412 MATHEMATICAL ANALYSIS [XVI, § 262 EXERCISES By use of the remainder theorem find the remainder when 1. x^ — 2 x'^ + 3 is divided by a; — 2. 2. a:i3 _ 45 a:i2 + 26 x^ + 12 is divided by a; - 1. 3. a:i2 + 1 is divided by jc + l ; by a; — 1. 4. Show that — 2 is a root of the equation 2 x^ + 3 a;^ — 4a; — 4 = 0. 5. Show that a;'* + «" is divisible by x + a if w is odd. 6. Show that x" + a" is not divisible by x + a if n is even. 7. By means of the remainder theorem find k so that x^-\- kx^-\-2x +^ may be exactly divisible by a; — 2. 8. Find the polynomial in x of the second degree which vanishes when x= I and when a; = 2, and which assumes the value 10 when x = 3. 263. Synthetic Division. We shall proceed to explain a simple method of performing the division transformation when the divisor has the form x — k, i.e. when the divisor is a binomial of the first degree in which the leading coefficient is 1. Let the given polynomial be a„aj" + «„-iaJ"~^ + a^-2X"'-^ + ••• + OjX + a^ and the divisor x — k. The ordinary process of long division leads to X — k a,.a;^— a„ A;.r"-i ^«^"~^ + (^«n + ^n-i)^""^ (ka, + ft^ -Qa;"-^ - k(ka^ + a.,-i)x^-^ It is now not difficult to see that (a) the first coefficient in the quotient is a„, i.e. the coeffi- cient of the leading term in the dividend ; (b) the second coefficient in the quotient is obtained by multiplying the first coefficient of the quotient by k and adding to it the second coefficient of the dividend : XVI, § 263] ALGEBRAIC METHODS 413 (c) the third coefficient of the quotient is obtained by mul- tiplying the second coefficient of the quotient by k and adding to it the second coefficient of the dividend. We may then arrange the work as follows : a„ a„_i a„_2 «„-3 - 1^ or„ ka^ -f a„_i k{ka„ + a„_i) + a„_2 - Here the first line contains the coefficients of the dividend in order and the third line gives the coefficients of the quotient and the remainder in order. Every number in the third line, after the first, is obtained by multiplying the preceding num- ber by k and adding to it the next number in the first line. Example 1. By synthetic division, find the quotient and the re- mainder when X*— 2ic3-f-x2 + 3a;— 2is divided by cc + 2. Solution : 1-21 3—21 — 2 - 2 8 - 18 30 1 _ 4 9 - 15 28 Hence the quotient is x^ — 4 ic^ + 9 x — 15 and the remainder is 28. Example 2. If/(x) = 2x* +3x3+7x2-}- 14x + 20, find/(-3). 2 3 7 14 201-3 -6 9 -48 102 2 -3 16 -34 |122=/(-3). Ans. EXERCISES In the following exercises use synthetic division : 1. Find the remainder when x^ + 3 x^ — 6 x + 2 is divided by x — 2. 2. Find the remainder and the quotient when x* — 3x'^ + 2x + 3 is divided by x + 3. 3. Show that 3 is a root of the equation x^ - 4 x^ - 17 x + 60 = 0. 4. Find k so that 3 is a root of the equation x'* - 3 x^ + A:x + 1 = 0. 5. Is 5 a root of the equation x^ — x^ + 7 = ? 414 MATHEMATICAL ANALYSIS [XVI, § 264 264. Properties of Integers. We have already noticed (ftn.j p. 403) that the familiar method of writing an integer greater than 9 represents it as the value of a polynomial. Integers and polynomials, therefore, have many properties in common. We may, for example, gain an insight into the reason for the rules of arithmetic used in adding a column of figures or in finding the product of two integers by comparing these rules with the technique of adding and multiplying polynomials discussed in § 256 *. We shall proceed to discuss some of the properties of integers relating to divisibility, etc., which are valuable in our everyday use of numbers. 265. Prime and Composite Numbers. An integer greater than 1 that is divisible by no integer except itself and 1 is called a prime number or simply a prime. Thus 2, 3, 5, 7, 13 are prime numbers. Any integer (> 1) not a prime is called a composite number. Any composite number is the product of two or more primes, thus 6 = 2-3, 100 = 2 • 2 • 5 . 5 = 22 . 5^. Any composite number 7i may be resolved into its prime factors in one and only one way. When resolved it has the form n = j?i*ij92"* •••Pfc"*- When a number has been resolved into its prime factors any question regarding its divisibility is readily answered by the following theorem. A number a is divisible by a number b if and only if every prime factor of b occurs in a to at least as high a power as it occurs in b. This theorem follows readily from exercises 15 and 17 below. The proof is left to the student. As an illustration, if a = 2 • 3^ • 17^ • 37 and 6 = 2 • 32 . 17 we recognize at once that a is divisible by b and that the quo- tient is 3 • 17 • 37. If, on the other hand, b were 2^ . 3^ • 17 then a would not be divisible by &. The common factors of two integers are also readily found if the num- bers have been resolved into their prime factors. Why ? Two integers which have no common factor (> 1) are said to be prime to each other. The notion of prime numbers and the investigation of their properties is very ancient and to this day some of the most difficult problems of advanced mathematics relate to this field. Some of the properties are quite elementary, however, and have been listed below in exercises. * Carrying this comparison out in detail forms a valuable exercise. The familiar process of "carrying" a digit from one column to the next is about the only thing that requires special investigation. XVI, § 265] ALGEBRAIC METHODS 415 EXERCISES 1. Prove that if a and h are both divisible by w, then a-\-h and a—b are divisible by n and a • 6 is divisible by n^. Is a similar theorem true of polynomials ? 2. Prove that a product of any number of integers is divisible by n if one of the integers is divisible by n. Is a similar theorem true of poly- nomials ? 3. If c = a & is divisible by «, must either a or 6 be divisible by n ? 4. Prove that if a number a leaves a remainder r when divided by &, then the number obtained by adding to a any multiple of h will leave the same remainder. 5. Prove that if the last digit on the right of an integer is even, the integer is divisible by 2. 6. Prove that if the number formed by the last two digits of an integer is divisible by 4, then the number is divisible by 4. 7. Prove that if the number formed by the last three digits of an integer is divisible by 8, then the integer is divisible by 8. 8. Prove that if the last digit of an integer is or 5 then the integer is divisible by 5. 9. Prove that if the sum of the digits of an integer is divisible by 3 (or 9) then the integer is divisible by 3 (or 9). 10. Prove that if the sum of the first, third, fifth, etc. digits of an integer is equal to the sum of the second, fourth, etc., the number is divisible by 11. 11. If the sum of the digits of an even number is divisible by 3, the number itself is divisible by 6. 12. Determine without performing the division whether the following numbers are divisible by 2, 3, 4, 5, 6, 8, 9, 11. (a) 2562. (c) 123453. (e) 127898712. {g) 111111111111. (6) 12345. {d) 2673. (/) 7325 x 8931. Qi) 11111111112. 13. How would you recognize that a number is divisible by 45 ? 14. Prove that if the product « • & is divisible by a prime number p, either a is divisible by p or 6 is divisible by p. Is a similar theorem true for polynomials ? 15. Prove that if a number c is a factor of ah and is prime to a, it must be a factor of h. Is a similar theorem true for polynomials ? 16. Prove that the quotients of two numbers by their H. C. F. are two numbers prime to each other. Is a similar theorem true for polynomials ? 416 MATHEMATICAL ANALYSIS XVI, § 265 17. Show that if a number is divisible by each of two numbers which are prime to each other, it is divisible by their product. Is a similar theo- rem true for polynomials ? 18. Show that the product of two numbers is equal to the product of their H. C. F. by their L. C. M. Is a similar theorem true of polynomials ? 19. Prove that the number of primes is unlimited. [Hint. Suppose that the theorem were not true and that p were the greatest prime. Let pi, /)2, P3, •••, Pn-i be the set of all primes less than p and consider the number PiP'zPz ■■' Pn-iP-^ 1- This number is not divisible by any of the primes pi, p2, •••, p. The rest of the proof should be obvious. This proof was first given by Euclid.] 20. By trying successive primes 2, 3, 5, 7, •••, determine whetlier or not 1009 and 1007 are primes. In this case we may stop with the prime 31. Wiiy ? Ans. 1009 is prime. • 21. Resolve into prime factors the numbers 604800 and 12250. 22. Is the number 2^31253 a perfect square ? Is it a perfect cube ? 23. Show that the relation ah — cd = 1, where a, &, c, rf, represent in- tegers, is not possible unless a and c are prime to each other. 24. Two consecutive integers are always prime to each other. Is this true also of any two numbers differing by 7 ? 25. What is the smallest cube of masonry that can be constructed of bricks 8x3x2 inches ? It is assumed that the bricks are placed so that any two equal sides are parallel. 266. Partial Fractions, in certain problems it is sometimes found necessary to express a fraction in which the numerator and denominator are polynomials in one variable as the sum of several fractions each of which has a linear or at most a quadratic function in the denominator. In what follows it will always be assumed that the given fraction is a proper fraction^ i.e. a fraction in which the degree of the numerator is less than that of the denominator. Any fraction which is not proper can be written as the sum of a polynomial and a proper fraction. Therefore our problem may be stated as follows : To express a proper fraction as the sum of several proper fractions. The method of approach is to assume that the fraction can be expressed in the desired form and then seek to determine the numerators which in the assumed form are left undetermined. Four distinct cases arise. XVI, § 266] ALGEBRAIC METHODS 417 Case I. When the denominator can be resolved into factors of the first degree all of which are real and distinct. Example 1. Resolve into partial fractions 9x^ -a;-2 x^ — X The sum of the three fractions X X —\ X -\-l will give a fraction whose denominator is x^ — x. We therefore try tc determine A^ B, C so that (Q) 9x^-x-2 _A B C X^ — X X X — 1 x -\- 1 Clearing of fractions we have (7) 9x-2-x-2 = A{x^ - 1) + B{x^ ■{■ x) -{- C {x'^ - x) . Since (7) is to be true for all values of x, we seek values of A, B, O, such that the coefficients of like powers of x will be equal, i.e. such that A + B-\- C^9, B-C = -l, -A =-2. Solving these equations, we find A = 2, B = S, (7 = 4. Hence 9a;2-a;-2_2 3 , 4 X^ — X x X—1 X -\- I Case II. When the denominator can be resolved into real linear fac- tors some of which are repeated. Example 2. Resolve into partial fractions 2 a;2 - X + 2 x(x-iy The sum of the fractions A^ B , C X x-\ {X- 1)2 will give a fraction whose denominator is x{x — l)^. Therefore we shall try to determine A, B, C so that 2x^-x + 2^A^ B , C X{X - 1)2 X X-l {X- 1)2 Clearing of fractions, we have 2x2-3-4-2 = A{x- 1)2 J^ Bx{x- 1)+ Cx. Equating the coefficients of like powers of x, we have A-^B = 2, -2A-B-\-C = -\, A = 2. 2e 418 MATHEMATICAL ANALYSIS [XVI, § 266 Solving for A, B, O, we find ^ = 2, B = 0, C = 3. Hence 2x2-a;+2_2 3 x(a'-l)2 X, (a;-l)2 The assumptions to be made under Cases I and II are contained in the following rules. (1) For every unrepealed factor x— a of the denominator^ assume a fraction of the form A/{x — a). (2) For every repeated factor {x — a)* of the denominator^ assume a sum of fractions of the form A\ ^ A2 _^ ,,, I Ak X — a (x— a)2 (x — ay Case III. W?ien the denominator contains quadratic factors which are not repeated. Example 3. Resolve into partial fractions 5 x2 + 4 X + 3 (X+1)(X2 + 1)' Let 5 x2 + 4 X + 3 ^ A Bx + G (x+ l)(x2+ 1) x + 1 x-^+ 1 ' Clearing of fractions, we have 5 a;-2 + 4 X + 3 = A(x^ +1) -{-(Bx+ C)(x + 1) = Ax^ + A -{- Bx'^ -^ Bx + Cx + a Equating coefficients, A-^B=o, B+C = 4,A+C=S. Therefore A = 2, B = S, 0=1. Hence 5x^ + 4x+3 ^ 2 . 3x4-1 (x+l)(x2 + l) x+1 X24-1' Case IV. When the denominator contains quadratic factors which are repeated. Example. Resolve into partial fractions 3 x^ + x'^ + 8 x'-! + X + 2 X(X2 + 1)2 Let 3 x^ + x^ + 8 x2 + a; + 2 ^A Bx + C Dx + E x(x2 +1)2 X x2 + 1 (x2 -f 1)^' Then, 3 x* + x8 + 8 x2 + X + 2 = .4(x2 + l)a + (^x + C)x(x2 + 1) + {Dx + E)x = ^x* + 2 ^x2 + A + Bx^ + Cx3 + ^x2 + Cx + Z)x2 + Ex. XVI, § 266] ALGEBRAIC METHODS 419 Equating coefi&cients we have ^ + JB = 3, C=l, 2^ + 5 + i) = 8, C+E = l, A = 2. Hence, A = 2, B = 1, 0=1, D = 3, E = 0. Therefore, ^^^ ^ x^ + Sx'^ + x + 2 ^2 x±l_ Sx x(x^+iy^ X a:2 -f 1 (a;2 + 1)2 The assumptions to be made in Cases III and IV are contained in the following rules. 3. Corresponding to every unrepealed factor of the form ax'^ + hx -\- c, assume the partial fraction Ax A- B ax?- + 5x + c 4. Corresponding to evet^ factor (ax^ -\- bx + c)*, assume the sum Aix + Bi J A2X + B2 _|_ ^^^ AkX 4- Bk ax^ -^ bx -\- c (aa;2 -f 6x + c)2 (q[x2 + 6a: + c)*' EXERCISES Resolve into partial fractions each of the following fractions. J 4x+ 1 jj 3 a:2 - 5 X + 4 (a;_l)(x+l)(x + 3) 3a;-l x2-4 ' 2x + l x2(x — 4) x^ + 1 X(X- 1)8* 1 x%x + 1) * x2 + 2 a; + 1 x^ + x 2 x2 - 1 3x3 + 3 x' 2x + l X3 + X2 + X * 1 12. 13. 14. 16. 16. 17. 18. 19 X(X2 + 1)2 3 X8-1' '" X2(X2+1)2 10. -^. 20 (X - 1)3 X2 (X2-1)2' X* (x2-l)(x + 2) x-2 (X + 1)X2 2 x2 - X + 3 x(x2-l)(2x-3) 1 +x^ (2 - x2) (2 + x2) ' X* + X2 X* + X2 + 1 3-2x2 (2 - 3 X + x2)2 5 x3 4- 2 X + 1 (X^+1)(X-1)2° 2x+l CHAPTER XVII PERMUTATIONS, COMBINATIONS, AND PROBABILITY THE BINOMIAL THEOREM 267. Definitions. Suppose that a group of n objects is given. Any set of r (r ^ n) of these objects, considered with- out regard to order, is called a combination of the n objects taJce7i r at a time. We often denote the objects in question, which may be of any kind, by letters, as a, 6, c, •••, k. The number of combinations of these n letters taken r at a time is denoted by the symbol „C^. For example, the combinations two at a time of the four letters a, b, c, d are, ab, ac, ad, be, bd, cd. Since there are 6 of these combinations in all, we have 4C2 = 6. On the other hand, any arrangement of r of these n objects in a definite order in a row is called a permutation of the n objects taken r at a time. The symbol „P^ is used to denote the number of such permutations. For example, the permutations of the four letters a, b, c, d taken two at a time are ab ac ad be bd cd ba ea da cb db dc. Since there are 12 of these arrangements in all we have 4P2 = 12. We have assumed in these examples that the objects are all different, and that the repetition of a letter within a permutation is not allowed. 268. Fundamental Principle. If a certain thing can be done in m different ways and if, when it has been done, a cer- tain other thing can be done in p different ways, then both 420 XVII, §269] PERMUTATIONS AND COMBINATIONS 421 things can be done in the order stated in m x p different ways. For, corresponding to the first way of doing the first thing, there are p different ways of doing the second thing ; corresponding to the second way of doing the first thing there are p different ways of doing the second thing ; and so on for each of the m different ways of doing the first thing. There- fore there are m x p different ways of doing both things in the order stated. This fundamental principle may at once be extended to the following form. If one thing can be done in m ivays, and if, when it has been done, a second can be done in p loays, and if when that has been done, a third can be done in q ways, and so forth, then the number of ways ill which they can all be done, taking them in the order stated, is m X p xq •••. Example 1. There are five trails leading to the top of Mt. Moosilauke, N. H. In how many ways may I go to the top, and return by a different trail ? There are five ways I may go to the top and for each of these there are four ways I may descend. Therefore, the total number of ways in which I may make the round trip is 5 x 4 or 20. Example 2. How many even numbers of two unlike digits can be formed with the digits 1, 2, 3, 4, 5, 6, 7, 8, 9 ? The digit in the units' place can be chosen in any one of 4 ways and the one in the tens' place can then be chosen in 8 ways. Tlierefore, 4 x 8 or 32 even numbers with two unlike digits can be formed from the given digits. 269. The Number of Permutations of n Different Things Taken r at a Time. The problem of finding the number of permutations of n different things taken r at a time can be stated as follows : Find the number of ways in which we can fill r places when we have n different things at our disposal. The first place can be filled in n ways, for we may take any one of the n things at our disposal. The second place can 422 MATHEMATICAL ANALYSIS [XVII, § 269 then be filled in ?i — 1 ways, and hence the first and second places together can be filled in n {n — 1) ways. Why ? When the first two places are filled, the third can be filled in w — 2 ways. Reasoning as before, we have that the first three places can be filled in n(n — l)(n — 2) ways. Proceeding thus, we see that the number of ways in which r places can be filled is n (ii — l)(?i — 2) ..• to r factors, and the rth factor is n — (r — l) or n — r + 1. Therefore the num- ber of permutations of n different things taken r at a time is (1) ^p^ = „(„_i)(n-2)...(n-r + l). Corollary. If r = n, we have (2) ^Pn = n (n - 1) (n - 2) ... 3 . 2 . 1 = n ! * Example. Three students enter an oflBce in which there are five vacant chairs. In how many ways can they be seated ? Here n = 6, r = 3. Hence sPs = 6 • 4 • 3 = 60 ways. 270. The Permutations of n Things not all Different. The number N of permutations of n things taken all at a time, of which p are alike, q others are alike, r others alike, and so on, is (3) N= ~ ■' Suppose the n things are letters and that p of them are a, q of them h, r of them c, and so on. Now, if in any of the N permutations we replace the p a's by p new letters, different from each other and also from the remaining n — p letters, then by permuting these p letters among themselves without changing the position of any of the other letters we can form p ! new permutations. Therefore if this were done in each of the N permutations, we should * The product of all the integers from 1 to n is called factorial n, and is de- noted by the symbol n ! or [n. Thus 3 ! = 1 • 2 • 3 = 6. A table of the values of n ! up to n = 10 will be found at the end of the book. XVII, § 270] PERMUTATIONS AND COMBINATIONS 423 obtain N-pl new permutations. In the same manner, if we replace the q 6's by q new letters differing from each other and the remaining n — q letters, the r c's by r new letters differing from each other and from the remaining n — r letters, and so on, we then obtain N -plqlrl "> new permutations. But the things are now all different and may be permuted in n I ways. Therefore J^ - p\ - q I - r \ "■ = nl, or p Iqlrl '" Example. How many different permutations of the letters of the word Mississippi can be formed taking the letters all together ? We have 11 letters of which 4 are s, 2 are p, 4 are i. Therefore the number of permutations is 11 !/(4 ! 4 ! 2 !) = 34650. EXERCISES 1. If there are six letter boxes, in how many ways can two letters be posted if they are not both posted in the same box ? Ans. 30. 2. If there are six letter boxes, in how many ways can two letters be posted ? Ans. 36. 3. Two dice are thrown on a table. In how many ways can they fall? Ans. 36. 4. Two coins are tossed on a table. In how many ways can they fall ? 5. In how many ways can five coins fall on a table ? 6. How many different permutations can be formed by taking five of the letters of the word compare ? 7. Find the number of permutations that can be made from all the letters of the word (a) assassination; (b) institutions; (c) examination. 8. Given the digits 1, 2, 3, 4, 5, 6, 7, 8, 9. Find (a) How many odd numbers of two digits each can be formed, repeti- tion of digits being allowed. (b) The same as (a), except that repetition of digits is not allowed. 9. How many even numbers less than 1000 can be formed with the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, repetition of digits not being allowed ? 10. In how many ways can a hand of ten cards be played one card at a time ? 11. In how many ways can 3 different algebras and 4 different geom- etries be arranged on a shelf so that the algebras are together? 424 MATHEMATICAL ANALYSIS [XVII, § 271 271. The Number of Combinations of n Different Things Taken r at a Time. TJie number of combinations of n different things taken r at a time is (4) „C, = "("-^)("-^)-("-'- + ^). r ! For, each combination consists of a group of r different things which can be arranged among themselves in r ! ways. Therefore ^C^ • r I is equal to the number of permutations of n different things taken r at a time ; that is, „C^ • r ! = „P,, or ^ _ n(n — l)(n — 2) >» (n — r + 1) -^~ r! Corollary 1. The value of „(7, may be written in the form rl{n — r)l This follows immediately from (4) if we multiply numer- ator and denominator by (n — r) !, since n(n- l)(n - 2) ... (w - r + 1) • (n - r) ! = n!. Corollary 2. The number of combinations of n different things taken r at a time is equal to the number of combina- tions of n different things taken (n — r) at a time. = „a. (n — r)l (??, — [n — r]) ! (n —r) I r\ The total number of ways in which a selection of some or all can be made from n different things is 2" — 1. For each thing may be disposed of in two ways, i.e. it may be taken or it may be left. Since there are n things, they may all be disposed of in 2" ways. But among these 2" ways is included the case in which all are rejected. Therefore the number of ways of making the selection is 2" — 1. XVII, § 271J PERMUTATIONS AND COMBINATIONS 425 Example 1. In how many ways can a committee of 9 be chosen from 12 people ? The required number is ^ _ ^ _ 12 . 11 . 10 _ „„^ 12t^9 — 12^3 — — q — r — — — ^IM. Example 2. From 6 men and 5 women, how many committees of 8 each can be formed when the committee contains (1) exactly 3 women ? (2) at least two women ? (1) The men may be chosen in eCs ways, the women in gCs ways. The number of ways in wliich both groups may be chosen together is eCs-eCs, or60. (2) Since each committee is to contain at least three women, it can be made up as follows : , . r ■, ^ (a) 5 men and 3 women. (b) 4 men and 4 women. (c) 3 men and 5 women. Therefore the number of possible committees is « 6^6 X 6<73 + 6^4 X 6^4 + 6C3 X 5^5 = 165. EXERCISES 1. Find 10 Cg; 11 10; 100C99. 2. How many different committees of 6 men can be chosen from a group of 20 men ? 3. There are 20 points in a plane, of which no three are in a straight line. How many triangles may be formed each of which has three of these points for its vertices ? 4. How many planes may be determined by 25 points, no four of which are coplanar, if each of the planes is to contain three points ? 5. How many different committees, each consisting of 5 republicans and 4 democrats, can be formed from 10 republicans and 8 democrats ? 6. From 20 men how many groups of 11 men each can be picked ? In how many of these groups will any given one of the 11 men be ? 7. Out of 6 different consonants and 4 different vowels, how many linear arrangements of letters each containing 4 consonants and 3 vowels can be formed? Ans. ^Ga x 403 x 7 !. 8. From ten books, in how many ways can a selection of six be made, (1) when a specified book is always included ? (2) when a specified book is always excluded ? 426 MATHEMATICAL ANALYSIS [XVII, § 272 272. Probability. If an event can happen in h ways and fail in / ways, and if each of these f+h ways is equally likely, the (mathematical) probability * of the event happening is h h-\-f and the probability of its failing is f/{h +/). An equivalent way of stating that h/(Ji -\-f) is the probability of an event happening is to say that the odds are h to / in favor of the event or / to 7i against the event. The probability of an event happening plus the probability of its failing is always equal to unity. Example 1. Suppose from a bag containing 3 red balls and 5 black ones, a ball is drawn at random, then the probability of its being red is f and of its being black f . The chance that the ball is either red or black is I + f = 1, or certainty. Example 2. From a bag containing 3 red balls and 6 black ones, two balls are drawn. Find the probability that (1) both are red, (2) both are black, (3) one is red and one is black. Two balls can be drawn in g C^ or 28 ways. Two red balls can be drawn in 3C2 or 3 ways. Therefore the probability of drawing two red balls is 3/28. Two black balls can be drawn in 5 (7-2 or 10 ways. Therefore the prob- ability of drawing two black balls is 10/28. The number of ways of drawing one red ball and one black one is zCi X 6^1, or 15. Therefore the probability of drawing a red and a black ball is 15/28. Example 3. Find the probability of throwing six with two dice. The total number of ways in which two dice can fall is 6 x 6 or 36. A throw of 6 can be made as follows: 1, 6; 5,1; 4,2; 2,4; 3,3; i.e. in 5 ways. Therefore the probability is 5/36. * The reason for the definition of mathematical probability may be made clear from the following considerations. Suppose a coin were tossed vi times and fell heads h times and tails/ times. If n is a finite number, h and/ will in general not be equal. But as n is increased, h/{h+f) and f/(h-\-f) will approach nearer and nearer to 1/2, and thus we take 1/2 to be the probability of the coin falling heads XVII, § 272] PROBABILITY 427 EXERCISES 1. In a single throw with one die, find the probability of throwing an ace. 2. In a single throw with two dice, find the probability of throwing a total of five ; six ; seven ; eight. 3. In a single throw with two dice, find the probability of throwing at least five ; six ; seven ; eight. 4. A bag contains 5 red balls, 6 green balls, 10 blue balls. Find the probability that, if 6 balls are drawn, they are (a) 2 red, 2 green, 2 blue; (b) 3 green, 3 blue; (c) 5 red, 1 green ; (d) 6 blue. 5. Four coins are tossed. Find the probability that they fall two heads and two tails. Ans. 3/8. 6. In a throw with two dice, which sura is more likely to be thrown, 6 or 9 ? 7. Find the probability of throwing doublets in a throw with two dice. 8. Five cards are drawn from a pack of 52. Find the probability that («) there is one pair. [Two like denominations make a pair, for ex- ample, two aces.] (6) Find the probability that there are three of a kind ; (c) two pairs ; (d) three of a kind and a pair ; (e) four of a kind ; (/) five cards of one suit. 9. Four cards are drawn from a pack of 52. Find the probability that they are one of each suit. 10. Seven boys stand in line. Find the probability that (a) a partic- ular boy will stand at an end; (6) two particular boys will be together ; (c) a particular boy will be in the middle. 11. A and B each throw two dice. If A throws 8, find the probability that B will throw a higher number. 12. Find the probability of throwing two 6's and one 5 in a single throw with three dice. * 13. In tossing three coins find the probability that at least two will be heads. 14. If the probability that I shall win a certain event is |, what are the odds in my favor ? 15. Find the probability of throwing an ace with a single throw of two dice. Ans. 11/36. 16. Which is more likely to happen, a throw of 4 with one die or a throw of 8 with two dice ? 428 MATHEMATICAL ANALYSIS [XVII, § 273 273. The Binomial Theorem for Positive Integral Ex- ponents. Consider the product (x -\-a)(x 4- a) ••• (x + a) [to n factors] where n is any positive integer. One term' of the product is oj" ; it is obtained by taking the letter x from each parenthesis. There will be n terms x^'^a, for the letter a may be chosen from any of the n parentheses which can be done in „0i = n ways. There will be „(72 terms x^'^a"^, for the a's may be chosen from two of the n parentheses and the x from the remaining n — 2 parentheses. In general, there will be nO^ terms a;""'" a**, for the a's may be chosen from any r of the n parentheses, and the x's from the remaining n — r parentheses. Therefore (6) (x + a)^ =zx^ + nCiX'»-ia + nCax'*-^ a^ + ... -\-„CrX'^-rar + ... + a". This formula for expanding (x -\- ay is known as the binomial theorem. Since „(7^ = ^(7„_„ it follows that the coefficients of any two terms equidistant from the beginning and the end are equal. If we write — a in place of a we have (x - ay = x-~' + nCiX--' (- a) + n^aaj'*'' (- ay + ... -f (- a)% or (aj-a)« = aj"— „Oia.-"-ia+ ^OgX^-^a^— ^CaOJ^-Vi^ -j 1- ( _ 1)" a\ Example 1. Expand (2x — yy. = 32 x5 - 80 x^y + 80 xV _ 40 x'Y + 10 a-y^ _ yS^ Example 2. Find the sixth term of (2 as — 3 yy. The sixth term is gCs (2xY ( - 3^)5, or - 108,864 x^. Example 3. Find what term contains x" in the expansion of ( x^ — J . Call it the «"i term. ThenioC«_i(x2)"-<^-iy~^ is the term. In this term we want the exponent of x to be 11. Therefore 22 — 2«— «-f-l = ll, or t = 4. The coefficient of this term is - 10C3 = — 120. XVII, § 273] BINOMIAL THEOREM 429 EXERCISES Expand the following by the binomial theorem : 1. (x-l)5. 3. {2x-yy. 5. (x--y, / l\io ^ ^' 2. (2x + y)6. 4- [^--) • 6. {z - xy)K 7. (0.9)6. [-Hint. 0.9 = 1-0.1.] 8. (0.99)3. Write down and simplify : 9.. The 8th term of (x - iy\ 12. The 6th term of (2 a; + 3 y)i2. 10. The 5th term of (2 x - y)io. 13- The middle term of (1 - xy\ U. The middle term of (2 x - y)i*. 15. The middle terms of {z - \/z) i^. 11. The 7th term Find the coefficient of \b ix) 16. x^ in the expansion of (x^ — -j , 17. a;i8 in the expansion of [x^-l--] . 1 \ 15 18. x^9 in the expansion of ( x* +- i . 19. x-^'' in the expansion of ( x* — ^ ] . V x^J 20. By considering the expansion of (1 + 1)», prove that 21. Prove 1 _ „Ci + „(72 - „C3 + •.- + (- 1)" „Cn = 0. MISCELLANEOUS EXERCISES 1. In how many ways can 10 boys stand in a row ? 2. In how many ways can ten boys stand in a row when (a) A given boy is always at a given end ? (6) A given boy is always at an end ? (c) Two given boys are always together ? (d) Two given boys are never together ? 3. How many numbers of three digits each can be formed from the digits 1, 2, 3, 4, 5, 6, 7, when (a) A repetition of digits is allowed ? (6) A repetition of digits is not allowed ? 430 MATHEMATICAL ANALYSIS [XVII, § 273 4. How many numbers of three digits each can be formed with the digits 2, 3, 5, 6, 7, 9, when (a) The numbers are less than 500 and a repetition of digits is allowed ? (&) The numbers are greater than 500 and a repetition of digits is not allowed ? 5. In how many ways can a consonant and a vowel be chosen from the letters of the word vowels ? 6. Find n when, (a)„(72 = 45; (6) „P3 .= 210 ; (c)„(72 = „C3. * 7. Show that the number of ways in which n things can be arranged around a circle is (n — 3) ! . 8. In how many ways can 6 people sit around a round table ? 9. How many signals can be made by hoisting 7 flags all at a time one above the other, if 2 are blue, 3 are white, and the rest are green ? 10. How many different numbers of seven digits each can be formed with the digits 1, 2, 3, 4, 3, 2, 1, the second, fourth, and sixth digits being even? 11. How many handshakes may be exchanged among a party of 10 students if no two students shake hands with each other more than once ? 12. A lodge has 50 members of whom 6 are physicians. In how many ways can a committee of 10 be chosen so as to contain at least 3 physicians ? 13. A crew contains eight men ; of these three can row only on the port side and two only on the starboard side. In how many ways can the crew be seated ? 14. Find n when „+2C4 = llnC2- 15. In how many ways can 18 books be divided into two groups of 6 and 12 respectively ? Atis. isC'e. 16. In how many ways can 12 students be divided into three groups of 4, 3, 5, respectively ? 17. How many different amounts can be weighed with 1, 2, 4, 8, and 16 gram weights ? 18. How many sums of money can be made with 5 one-cent pieces, 4 dimes, 2 half dollars, and 1 five-dollar bill ? 19. In how many ways can four gentlemen and four ladies sit around a table so that no two gentlemen are adjacent ? Ans. 144. XVII, § 273] PERMUTATIONS AND COMBINATIONS 431 20. Prove nO, + „Or_l = n+lC,-* 21. How many dominos are there in a set numbered from double blank to double six ? 22. A railway signal has three arms and each arm can take three different positions. How many signals can be formed ? 23. Prove n+^Cr+i = nCr+i + 2 „0, + „a-i. 24. How many combinations of four letters each can be made from the letters of the word proportion ? How many permutations ? Ans. 53; 758. 25. Find the probability that in a whist hand a player will hold the four aces. 26. Find the probability of drawing a face card from a pack of 52 playing cards. 27. If two tickets are drawn from a package of 15 marked 1, 2, •••, 15, what is the probability that they will both be marked with odd numbers ? both with even numbers ? both with numbers less than 10 ? both with numbers more than 10 ? 28. To decide on partners in a game of tennis four players toss their rackets. The 2 "smooths" and the 2 "roughs" are to be partners. What are the odds against the choice being made on the first throw ? 29. Prove that the sum of the coefficients of the odd terms of a binomial expansion equals the sum of the coefficients of the even terms. 30. If n is an even integer, prove that there is a middle term in the expression of (x + a)" and that its coefficient is even. 31. Provethat„Ci+2„(72 + 3„03 + ••• n„a„= w(2)"-i. 32. Prove „Ci - 2„C3 + 3„(73 + .- (- l)"-i • w • „C„ = 0. * An application of this formula is the construction of Pascal's Triangle. (o^o by definition will be assigned the value 1.) oCo iCo iCi 2C0 aC'i . 2C2 8^0 sCi 8C2 3^3 4C0 4C1 4C2 4C3 1 2 1 3 3 1 4 6 4 404 The formula in Ex. 20 shows that any number n+iCr is equal to the number just above it, i.e. nCr, plus the number nCr-i which is to the left of nCr.. Thus for example 4C8 = 3C3 -f 362. We can, by means of this formula in Ex. 20, write down the next row. It is 1 5 10 10 5 1 The numbers in the nth row of the table are seen to be the coefficients of the terms in the expansion of (z-{- a)^ (§ 273) . CHAPTER XVIII COMPLEX NUMBERS 274. Definitions. We have already had occasions to refer to the so-called imaginary numbers. A number that arises as the result of extracting the square root or, indeed, any even root of a negative number is called an imaginary number. Thus V— 2 is an imaginary, number ; the roots of the quadratic equation ^624. 3 _ q^ yiz. ± 2V— 2, are imaginary numbers. We have hitherto avoided the use of imaginary numbers as far as possible. It now becomes desirable to take them defi- nitely into account, to learn how to work with them, and to gain some knowledge of their usefulness. Indeed, one of the primary objects of this chapter is to show that imaginary numbers have quite as concrete an interpretation as the real numbers, an interpretation which in many cases is of great service in the solution of concrete problems. The letter i is used to represent the so-called imaginary unit; it is by definition such that i^ = — 1. Numbers of the form ibj where 6 is a real number different from zero, are called pure imaginary numbers. Numbers of the form a -\- ib, where a and b are real numbers, are called complex numbers. In the complex number a -f- ib, a is called the real part and ib the imaginary part. In a real number the imaginary part is zero ; in a pure imaginary the real part is zero. A complex number a -f- &« is imaginary if 6 ^ 0. When two complex numbers differ only in the sign of the imaginary part they are said to be conjugate. Thus 3 + 2 t and 3 — 2 i are conjugate complex numbers. 432 XVIII, § 276] COMPLEX NUMBERS 433 275. Assumption. We assume that complex numbers obey the laws of algebra given in § 41. By applying this assump- tion we have symbolically for the sum and difference of the two complex numbers a + ib and c -f id, a -{-lb ± (c -\- id) = a ± c -\- i(b ± d). That is, to add (subtract) complex numbers, add (subtract) the real and imaginary parts separately. 276. The Geometric Interpretation of the Imaginary Unit. We now seek a geometric interpretation of the imaginary unit i. To this end we recall the familiar representation of the real numbers as directed segments on a line, o j together with the interpretation of multiplica- ^^co^-a a tion by - 1 (§ 35). To multiply a real number ^'''- ^^^ a by — 1 is equivalent geometrically to a rotation about the point through two right angles of the segment OA which represents a (Fig. 237). Now, by definition, i is such a number that i^ = — 1. To multiply a real number a by — 1 is then equivalent to multi- plying it by P, i.e. by i • i. Multiplying a real number a by i may, therefore, be interpreted geometrically as an operation which when performed twice is equivalent to a rotation about in the plane through two right angles ; i.e. to multiply a by i may be interpreted geo- metrically as equivalent to rotating OA about " ^ in the plane through one right angle. The number ai will then be represented by a segment OB equal in length to OA whose direction makes with that of OA an angle of 90° (see Fig. 238). In the figure we' have also indicated the result of multiplying ci by i^=i 'i=—l and by i^=i • i • i= —i. 2f 434 MATHEMATICAL ANALYSIS [XVIII, § 276 Multiplying by i* = i • i . i . i = 1*2 . /2 = 1 is then to be inter- preted as a rotation through four right angles. EXERCISES Give the conjugates of the following complex numbers : 1. 3 + 2 I. 2. 3 - 4 1. 3. _ 5 - 3 1. 4. _ 8 + i. Simplify the following expressions : 6. 2(3 + 4 i) - 4(1 - i). 7. ll^ii _ ^J^li. 6. - 4(1 - i) -}- 6(3 - 28 i). 8. x + iy + ix + y. 9. Prove that the sum of two conjugate complex numbers is a real number. 10. Is the following statement true ? If the sum of two complex numbers is a real number, the complex numbers are conjugates. Explain. 11. Prove that every even power of i is equal to either 1 or — 1. 12. Prove that every odd power of i is equal to either i or — i. 13. Find the value of i 4- 2 i^ + 3 i^ + 4 i*. 14. Find the value of i^^ + i*^ + ^es _|. j69 _,. 4-44. 277. Vectors in the Plane. We have seen that, if any real number a is represented by a horizontal segment directed to the right or left according as the number a is positive or negative, then the imaginary number ai may be represented by a vertical segment directed upward or downward accord- ing as a is positive or negative. This suggests the possi- bility of representing other complex numbers by segments having other directions in the plane. Such a directed seg- ment will represent a magnitude (the length of the segment) and a direction. Therefore such a segment can be used to represent a variety of concrete quantities that are not merely geometric; e.g. a force of a given magnitude and acting in a given direction; a velocity, meaning thereby the speed (magnitude) and the direction in which a body moves ; etc. Such quantities having both direction and magnitude are XVIII, §278] COMPLEX NUMBERS 435 called vectors, and, if the directions are restricted to lie in the same plane, they are called plane vectors. Any plane vector may, then, be represented by a directed line-segment in the plane. Two vectors are said to be equal if and only if they have the same magnitude and the same direction. Hence,- from any point in the plane as initial point, a vector can be drawn equal to any given vector in the plane. 278. Addition of Vectors. The addition of vectors in the plane proceeds according to a definition analogous to the geo- metric addition of directed line-segments discussed in § 35. If we are given two vectors AB and BC, we may conceive the first to represent a motion from Ato B and the second a motion from B to O. The sum of the two vectors then represents, ^ ^ by definition, the net result of moving from A ^^^' ^^^ to B and then from B to C, i.e. the motion from A to C. The sum of the vectors AB and ^C is then the vector AC (Fig. 239). In symbols ^s + BO = AC. In other words, the sum of two vectors is the vector from the initial point of the first to the terminal point of the second, when the vectors are so placed that the initial point of the second coincides with the terminal point of the A,^^^^\/ first. From this definition it follows immediately ^ that, if two vectors issue from the same point 0, their sum is the diagonal, issuing from 0, of the parallelogram of which the two given vectors form two adjacent sides (Fig. 240).^ * If the vectors represent two forces, this shows that the sum of the vectors represents the resultant of the forces according to the law known as " the parallelogram of forces." 436 MATHEMATICAL ANALYSIS [XVIII, § 279 279. The Components of a Vector. The projection of a vector on a given line is called its component parallel to the line. Thus in Fig. 241 the directed segment M1M2 is the horizontal »« r -^ ' *l 1 i 1 M, -M, Fig. 241 component of the vector AB, and the directed segment ^1:^2 is its vertical component. Moreover, vector AB — vector N^N^ + vector M^M^. If the horizontal and the vertical components of a vector are known, then the vector is known. Why ? 280. The Complex Number x + iy and the Points in the Plane. Let OP (Fig. 242) be any vector issuing from 0, and let the horizontal vectors issuing from be represented by the positive and negative real numbers (and zero). We have seen F; p Vi ■^ X J / V Fig. 242 that the numbers of the form ai can be represented by the ver- tical segments issuing from 0. Here a is a real number and i is a vector of unit length. The horizontal component of OP will then be a certain real number x, and the vertical component a certain pure imaginary number iy. The vector OP will then be equal to the sum of these two components, i.e. OP^x-\- iy. XVIII, § 280] COMPLEX NUMBERS 437 Conversely, every number of the form x -\- iy represents a definite vector in the plane. If its initial point is at the origin of a system of rectangular coordinates (with equal units on the two axes), its terminal point is the point (x, y). We have hitherto used vectors in the plane to represent the complex numbers. If we think of these vectors as all having their initial points at 0, each vector determines uniquely, and is uniquely determined by, its terminal point. Hence, we can also use a complex number to represent a point in the plane, viz. the number x + iy will represent the point whose rectan- gular coordinates are (cc, y). Example 1. Represent by means of vectors the complex numbers '2 ■+ 2i and 1 ■\-Qi. Find the vector that represents their sura. In Fig. 243 the vector OA represents the complex number 2 + 2 i, and the vector OB represents the complex number 1+6 i. The sum of these two complex numbers is represented by the vector OC. Why ? -p -■e ^t -r, ^-!i ^^ 71 ' it _, , ' - ===f K-"-'- = : :::Si " - irM t -Z^ t 1 t it ^A t 4 l^X- -0- - ^ 1 _x T /-y > \\ \ \ \ "^ \ h ^i ^\ q2I. N Y ^v -^ \ A^ '' ^ B • - Fig. 243 Fig. 244 Example 2. Find the vector that represents (1 + *") — (2 — 3 1). To find this vector, find the vectors, OA and 0J5, that represent 1 + i, and 2 — 3i, and determine OC so that OA is the diagonal through O of the parallelogram of which OB and OCare adjacent sides (Fig. 244). Note that the vector OC is equal to the vector BA. 438 MATHEMATICAL ANALYSIS [XVIII, § 281 281. Equal Complex Numbers. Ifx-{-iy = 0, then x = and y = 0. For, if cc + iy = 0, and y=^0, we should have x/y = — i, which is impossible. Why ? If Xy + lyi = X2 + iy2, then x^ = x^ and y^ = 2/2- ^oi*) ^J trans- posing terms, we have (x^ — Xo) + i(yi — 2/2)= ^- Hence, we have Xi = X2 and y^ = 2/2. Thus, fwo complex numbers are equal if and only if the real part of the first is equal to the real part of the second, and the imaginary part of the first is equal to the imaginary part of the second. Geometrically, two complex numbers are equal if and only if they represent the same point. 282. The Polar Form of a Complex Number. Connect the point P{x, 2/) (Fig. 245), which represents the complex number X 4- iy, to the origin 0. If we let {p, 6) (p > 0) be the polar coordinates of P(0 being the origin and OX the initial line), then for any position of the point F we have ^ 'x = p cos 6, ,y = p sin 0. Therefore, the complex number x -\-iy may be written in the form (2) x + iy = p(cose + zsine). (p^O.) This form of complex number x + iy is called the polar form. The angle 6 is called the angle or the argument, and the length p is called the absolute value * of the complex number. Example. Find the angle, the absolute value, and the polar form of the complex number 2 -}- 1 2\/3. Plot the complex number (Fig. 246). Now we have p — y/x^ + y^. Hence p = \/4 + 12 = 4. Moreover tan^=\/3, i.e. 6=00°. That is, the absolute value is 4 and the angle is 60°. Therefore the polar form is 4 (cos 60° + i sin 60°) . ^«- 2*6 * Also sometimes called modulus. V (1) Fig. 245 Y J Pi2+i2/3) / 2/3 A'° 2 X XVIII, §282] COMPLEX NUMBERS 439 EXERCISES In the following exercises represent by vectors the numbers in paren- theses, and their sum or difference as the case may be : 1. (3 + + (-4 + 20. 4- (5-40-(-2-i). 2. (1 +3i)-(5-60. 5. (3 + 2z)+(3+20. 3. 7_(5 4.3 4). 6. (-4-4i)-6. Represent by a point each of the following complex numbers ; 7. 3 + 5 i. 9. 6 + i. 11.-3+6 i. 8. 3-3i. 10. -5-3i. 12. 7 + iV2. In the following exercises, represent by points the numbers in paren- theses, and their sum or difference as the case may be : 13. (3-hO + (-4 + 2 0. 16. (5-4 0-(-2-0- 14. (l-f-3 0-(5- 6i). 17. (3 + 2 + (3-t-2i). 15. 7 -(5 + 3?). 18. (3 + 3 0-5. Find real values of x and y satisfying the equations : 19. 2x— iy = 4:y — Q — 4i. 22. ixy + ic+?/ = 5 + 4i. 20. X + io-y = y + 6 + 36 i. 23. a:2 + ?/2= 25 - (3 x+4 y-2b) i. 21. {Sx + Qy+2)i — Sy-x=8. 24. ix -{- iy = 4: i -\- 5 x. Find the angle and the absolute value of each of the following complex numbers. Represent the numbers in polar form : 25. l + iV3. 27. 1-1. 29. 3i. 31. -8i. 26. 5 + 5i. 28 1-*-^. 30. -8. 32. 12 + 5 i. 2 2 33. Can the complex number x + iy, where x and y are real numbers, equal 7 ? 34. Under what circumstances is the sum of two complex numbers a real number ? Change the following complex numbers from the polar form to the form x-}-iy : 35. 3(cos 30° + i sin 30°) . 38. 2 V2(cos 225° + i sin 225°) . 36. 4(cos 135° + z sin 135°). 39. 4(cos 90° + i sin 90°). 37. cos 210° + i sin 210°. 40. 8(cos 180° + i sin 180°). 440 MATHEMATICAL ANALYSIS [XVIII, § 283 283. Multiplication of Complex Numbers. Our assump- tion in § 275 allows us to multiply two complex numbers iCi -f- iz/i and X2 + iy^ as follows : (a?! + *2/i)(aJ2 + m) = 3^1352 4- iy 1X2 + ixiy2 + ^^l2/2 = (aJia^2 - 2/1^2) + *(^i2/2 + X2yi)' If the two numbers are written in polar form, the multipli- cation may be performed as follows : ^'1 + *2/i = pi(cos 61 4- i sin ^1), ^2 + %2 = p2(cos $2 + I sin ^2)- By actual multiplication, we have (•^1 + iyi)ix2 + m) = P1P2 [cos 61 cos ^2 -h '' (sin Oi cos ^2 + cos Oi sin ^2) — sin Oi sin ^2] = P1P2 [cos (^1 4- ^2) + i sin ((9i + ^2)].* Therefore, the absolute value of the product of two complex numbers is equal to the product of their absolute values, and the angle of the product is equal to the sum of their angles. In Fig. 247 the points Pi and P2 represent the complex numbers xi + iyi and X2 + iy2 respectively. The point P3 represents {xi-\-iyi)(x2 + iyo)- Fig. 247 284. Division of Complex Numbers. The quotient of two complex numbers Xi -\- iyi and X2 -f- iy^ luay be reduced to the form a + ib if we make the denominator real by multiplying both numerator and denominator by the conjugate of the denominator. Thus, Xi + iyi _ a?i + m . X2 — iy2 ^ X1X2 -h iyiXj — ixiy2 — i^yiyz X2 -f iyo «2 + m ^2 — iyi ^2' + 2/2' * See § i:ts X2^ + 2/2' ^%Vi x^^y^ XVIII, § 284] COMPLEX NUMBERS 441 If we write the two complex numbers in polar form and then perform the division, we have pi (cos $1 + I sin ^i) _ pi (cos Oi + i sin ^])(cos 0^ — ^ sin O^) P2 (cos $2 + i sin O2) p2 (cos O2 + i sin 62) (cos ^2 — i sin ^2) ^ Pi [cos (^1 - O2) + i sin (^1 - ^2)1 P2(cos2^2 + sin2|92) = ^^ [cos (^i - $2) + 1 sin (0, - ^2)]. Therefore, the absolute value of the quotient of two complex numbers is equal to the quotient of their absolute values, and the angle of the quotient is equal to the difference of their angles. Example 1. Find analytically and graphically the product (1 -f i) Solution. Analytically, (1 + 0(3 4. V8 i) = 3 + 3 I + V3 ^ + \/3 i^ = (3 - V3) + i (3 + V3). Graphically, writing the complex numbers in polar form, we have V2(cos 45° + i sin 45°) and 2 V3(cos 30° + 2 sin 30°). Therefore pi = \/2, P2 = 2 V3, Ox = 45°, 62 = 30°. Ir 4^ t^ £ 1 t -^rit - 7 p^^^' tc-^ ^_ X 1 . . Fig. 248 Hence the absolute value of the product is p\p2 =2 V6 and the angle of product is 75°. In Fig. 248 the points Pi, Pi, and P represent respectively the complex numbers 1 + i, 3 + iVS, (1 + 1)(3 + iV3). 442 MATHEMATICAL ANALYSIS [XVIII, § 284 Example 2. Find analytically and graphically the quotient (3 + iV3)/(l + 0- Solution : Analytically : 3 + iv/3 ^ 3 + iV3 1-i^ (3+ V3)-i(3-V3) ^ l+i 1 + t 1 - i z ^x c^ -t //^ gr ^_ jo^^ ^±: ■-E Fig. 249 Graphically^ using the results in Ex. 1, we see in Fig. 249 that the points Pi, P2, and P represent respectively the complex numbers (1 + 0, (3 + iV3), (3 + iV3)/(l + 0. EXERCISES Perform the following operations analytically and graphically : 1. (1 + 0(2 + 2 0. _ g l-zV3 2. (1 + i V3)(2 4-i2V3). * -3 ' 3. (2 0(5)- T 5 + 5t 4. (i_j-0(-2-20(- 1 + iV3). * l-» 3 + iV3 8. n-0' - 6. 1 + I 2 + I 2 \/3 Perform the following operations analytically : a 3 + i U. (i9 + iio + i" + ii2)7. l_^18t 3-29t 10. 7-iV2 3 + 4 i 3 - 4i 11. 12. 13. (2 + 1)2 (2 - 0' X + t Vl — X2 16. V V2 / Vl-x^ 17 2 + 3 ;; 3 + 2 ^ 3 _ 4 I 3 + 4 i 1 + i 18. V? + 24 i. XVIII, § 285] COMPLEX NUMBERS 443 285. DeMoivre*s Theorem. The result of § 283 when ap- plied to the product of any number of complex numbers leads to the following : I. The absolute value of the product of any number of complex numbers is equal to the product of their absolute values. II. Tlie angle of the product of any number of complex numbers is equal to the sum of their angles. If the above statements be applied to a positive integral power of a number, i.e. to the product of n equal factors, we obtain (3) [/3(cos 6 -\- i sin ^)]" = p"(cos nd -\- i sin n 6). For the special case p = 1 we obtain (4) (cos 6 + i sin 6)" = cos n 9 + / sin n 6. This relation we have just proved for the case where n is a positive integer. It also holds when ti is a negative integer. For we have (cos^ + isin^r= i ^cos^-tsin^ ^ cos + 2 sin e cos2 6 -f- sin^ 6 = cos (— 0) 4- i sin (— 0), and hence (cos + i sin 6)-'' = [cos (— 0) -f i sin (— 0)^ = cos (— p 6) 4- i sin ( — p 0). Further, \i n — 1/g, where g is a positive or negative integer, we have, by what precedes, (5) (cos + i sin Of = [Uo^- + i sin-Vl^ and hence (6) (cos ^ + i sin Oy = (^cos - -f- / sin ^\ = cos ^ ^ + / sin ^ 9. 6 13= COS-+ isin-, q q 444 MATHEMATICAL ANALYSIS [XVIII, § 285 This shows that relation (4) is valid for all rational values ofn. It should be noted, of course, that relation (5) states merely that a certain q^^ root of cos 6 + i sin 6 is cos {O/q) -f- i sin {6/q) and that a similar statement applies to relation (6). The fact expressed by (4) is known as De Moivre's theorem.* 286. "Powers and Roots of Numbers. De Moivre's theo- rem often enables us to compute an integral power of a complex number without difficulty, as the following example will show. Example 1. Find the value of (2-{-2i)^. The polar form of this number is 2 \/2(cos 46'^ + i sin 45") . Hence (2 + 2 0^= (2\/2)6(cos225° + i sin 225°) = 128V2(---- 7:^=- 128 -128 I. V V2 V2y To find the nth roots of a number requires special methods. Example 2. Find the 5th roots of 2 + 2 i. Here as in Ex. 1 we may write 2 + 2 i = 2V2(cos 45° + i sin 45°) and hence (2 +2 0^=(2V2)^(cos9° + isin9°). But this is not the only number whose fifth power is 2 +2 i. For we may write 2 + 2i = 2v'2[cos(45° + k 360°)+ i sin(45° + fc 360°)], where k is any integer. That is to say, (2 + 2i)^=(2\/2)^[cos(9°+ ^•72°)+ isin(9° + Ar72°)]. For the values A; = 0, 1, 2, 3, 4 we get the five numbers (7) (2\/2)^"(cos9° + isin9°), (2 V2)^ (cos 225° + i sin 225°), (2\/2)^^(cos81° + i sin 81°), (2V2)^(cos297° + i sin 297°). [ (2 V2)^(cos 153° + i sin 153°), The succeeding values of k (i.e. A = 5, 6, •••) evidently give numbers equal to the preceding respectively. Each of the five numbers is a fifth root of 2 + 2 i ; they are all different. ♦Abraham de Moivre (lfi67-1754), a mathematician of French descent who lived most of his life in England. XVIII, § 286] COMPLEX NUMBERS 445 The general formulation of the problem of finding the nth root of a number z = /o(cos 6 -\-i sin 9) is as follows. The most general form for z is z = p[Gos{e + k 360°) 4- i sin((9 + k 360°)], where k is an integer. This gives, by De Moivre's theorem, 1 1 z'-==p' (9 + A: 360° , . . 6> + A; 360 cos --^ h t sm —^ The n values A: = 0, 1, 2, —, n — 1 give n different values for 2^/" and no more values are possible. Why? Here pV« means the numerical nth. root of the positive number p. We have then: Every com2i>lex number (^0) has just n nth roots. These n roots all have the same absolute value ; their angles may be arranged in order in such a way that every two successive ones differ by 360°/7i. EXERCISES By using De Moivre's theorem find the indicated powers, roots, and products. 1. (4 4-i4V2)6. 4. (3 + iV3)io. 2. (cos 10° + i sin 10°)9. 5. ( - 1 - WS)^. 3. (l+iV^. 6. (-2+2«)*. 7. [3(cosl5° + isinl5°)]i5. 8. [2(cos 20° + I sin 20°)][3(cos 70° + i sin 70°)]. 9. [2 4-2i][V8 + t]. 10. (3-3 0(-l + iV3). 11. \/4 + i4>/2. 12- V3 + tV3. 13. </_4+4z. 15. V- l-iV3. 16. v^cos 45° + i sin 45^. 17. V2Ti. 18. The cube roots of 1. 19. V:^. 20. V2^. 14. \/8(cos6U° + isin60^). 21. Prove that tlie n nth roots of a given number z are represented by the vertices of a regular polygon of n sides whose center is at the origin. 446 MATHEMATICAL ANALYSIS [XVIII, § 287 287. Applications in Trigonometry. De Moivre's theorem may be used to advantage in certain trigonometric problems. I. To express cos nO and sin nd in terms of cos and sin 6. ' We have the relation cos nO + i sin nO = (cos $ -{- i sin Oy = cos" O-hn- i cos"-i ^ sin ^ -f ^K^^ — ^) ^-2 cos"-2 sin^ ^ + ... A If in this relation we equate the real and the imaginary parts we get the expressions desired. Example 1. Express cos 6 d and sin 6 ^ in terms of cos 6 and sin d. The above method yields in this case : cos 6 ^ + i sin 6 ^ := (cos 6 ■{■ i sin 0)'^ = cos6 e -\-Q icos^ ^sin ^ - 15 cos* 0sin2 ^ - 20 1 cos^ 6 sin' 6 + 16cos2 ^sin* d + 6 I cos 9 sin° 6 — sin^ d. Equating the real parts, we have cos 6 £> = coss e — 15 cos* e sin2^ + 15 cos'^ sin* d — sin^ 9. Equating the imaginary parts we get (after dividing by i) sin 6 ^ = 6 cos5 d^inO — 20 cos^ d sin^ ^ + 6 cos ^ sin^ d. II. To express cos"" 6 and sin"" 6 in terms of sines and cosines of multiples of 0. If we place w = cos + i sin 0, we have u^ = cos kO 4- i sin kO, u~^ = cos kO — i sin kO. Adding and subtracting these equations, we have .r,. [ w* + w'' = 2 cos k$, [ u^ — n -* = 2 4 sin kd, for any integral value of k. In particular when /c = 1, we have 2 cos 6 = u -^ u~^, 2 i sin $ = u — ^^~^ It follows that 2^ cos" e= {u -|-ifc-i)«= w"4- nu-'^-{- ^^^'~^\ ''-*-\ \-na-^''-^^+u'\ XVIII, § 287] COMPLEX NUMBERS 447 The fact that the coefficients in the binomial expansion are equal in pairs makes it always possible to group the terms as 2'^ cos"^ = {w H- ?r") -h n(ifc'^-2 -f ir(«-2))+ .... But the terms in parentheses on the right are equal respec- tively to 2 cos ?i^, 2 cos {n — 2)0, •••. The following examples will make the method clear. Example 2. Express cos* in terms of cosines of multiples of 6. We set = w4 + 4m2_|.6 + 4 u-^ + M-* = (m* + M-*) + 4(m2 _|_ u-2) -f 6 = 2 cos 4 ^ + 4 . 2 cos 2 + 6. Dividing both members by 2* we obtain the desired result cos* e = I (cos 4 ^ + 4 cos 2 ^ + 3) . Example 3. Express sin^ 6 in terms of multiples of the angle d. We set 25 1*5 sin^ e = {u— u-^y or 32 i sin^ 6 = u^ — bu^ -\-10 u— Id ir^ + 5 m-^ — u-^ = («5 _ |<-5) _ 5(1(3 _ «-3) 4. 10(w _ M-1) = 2 I sin 5 ^ - 5 • 2 i sin 3 ^ 4- 10 . 2 i sin d. Whence sinS d = ^^ (sin 5 ^ — 5 sin 3 ^ + 10 sin ^) . EXERCISES Express each of the following in terms of cos d and sin 0. 1. cos 2 and sin 2 ^. 3. cos 4 ^ and sin 4 ^. 2. cos 3 e and sin 3 ^. 4. cos 5 ^ and sin 5 0. 6. Show that tan 4 g ^ ^ <^^^ ^ (^ - ^^"' ^) . 1-6 tan2 + tan* ^ 6. Find tan 5 ^ in terms of tan 0. Express each of the following in terms of sines and cosines of multi- ples of ^ : 7. sin8 0. 9. sin* 0. 11. cos^ 0. 8. cos8^. 10. cos«^. 12. sin6^. 418 MATHEMATICAL ANALYSIS [XVIII, § 287 MISCELLANEOUS EXERCISES Solve the following equations and illustrate the results graphically. 1. x3 - 1 = 0. 3. x5 - 32 = 0. 5. cc8 - 1 = 0. 2. x3 + 1 = 0. 4. a;6 - 1 = 0. 6. a:^ + 1 = 0. 7. Prove that cos nd = I [cos d + I sin ^]" + ^[cos 6 — i sin ^J". 8. Prove that i sin nd = I [cos 6 -\- i sin ^]" — i [cos — i sin ^J**. 9. Prove that /I -f sin ^ + t cos ^X'* ,, /,N , • • /T „N { '—- — =cos(lmr—nd)+ismUn'ir — nd). 10. Prove that the product of the n nth roots of 1 is 1, if n is odd, and — 1 if n is even. 11. Prove that the sum of the n nth roots of any number is 0. 12. Complete the discussions in § 287 to derive the following formulas. L (a) cos;i^=cos"0— ^^^^i^~!-)cos"-'^^sin2^ ^ ^ 2! , n(n— l)(n — 2)(n — 3) „ a ^ ■ a ^ . _l — V ZA zs 1 cos "-4 d sm^ 6 -\- •••. 4 1 (6) sin nd=:n cos'»-i ^ sin ^ - ^(^ - ^) ( ^.ZL^ cos^-s q gin^ ^ o ! , nCn- l)(n - 2)(n-3)(n-4) „ ,^ . ,^ + ~^^ ^-^ — — ^ cos"-fi e sin^ ^4- . . •. 5 ! II. (a) cos«^=-l- rcosn^+ncos(n-2)^+*-^— ^cos(n-4)^+-"l. n (b) sin»^ =^-=^1 cosn^- ncos(n - 2)^ +^^^=-^cos (n-4)^+ •••], if n is even ; but sin"/? = i=J-) ^ rsinn^-nsinrn-2)^4- ^^^~-^^ sii if n is odd. fsin «^-n sin(n-2)^ + ^^^ ^^ sin (n - 4)^+ ...], CHAPTER XIX THE GENERAL POLYNOMIAL FUNCTION THE THEORY OF EQUATIONS 288. The General Polynomial Function of Degree n. The general polynomial of degree n, f{x) = a^x"" 4- a^-ix""-^ + a,^-2X^-^ H \- aiX -{- a^ (a„ :^ 0), has already been defined (§ 255). We have already dis- cussed in some detail special cases when the degree of f(x) is 1, 2, 3, (Chapters III, IV, V). For these cases we proved that the function is always continuous, and we learned how to find the slope of the graph of the function at any point. It is our present purpose to extend these results and methods to a func- tion represented by a polynomial of any degree. ax^ + a^-iX""-^ H h aiK + otr 289. The Slope of the Graph of /(x). the slope of the graph of the equation (1) y=n^) at any point Pi{xi, y^ of this graph, we first find the slope A^z/Aic of the secant P^Q (Fig. 250) joining the point Pi to any other point Q(flJi+ Aa?, 2/i + Ay) on the graph. To this end we must first calculate the value of Ay in terms of Xi and Aoj. We have 2o 449 Continuity. To find Fig. 250 450 MATHEMATICAL ANALYSIS [XIX, § 289 = a,(a^i + ^xy + a«_i(a:i + Axf-'^ + - +ai(xi + Ax)-\-aQ. Vi =/W= «„a;i" + a„_iaji«-i + .•• + a.x^ + a^. By subtraction and proper grouping of terms we find (2) Ay=f(x,-{.Ax)-f(x{) = alix, + Axy- a^r] + a„_i[(a;i + Axy-^-x^^^-] H h«i[(^i + Aa;) — ajj. Each of the terms of this expression is of the form (3) al(x, + Axy-x,'^l and the whole expression is equal to the sum of all terms ob- tained from (3) by letting k take on the values k = n,n^l, —, 1. Expanding the first term in the brackets, we obtain a,l{x^ + Axy - X,''] = «,[.T/+A;a;i*-iAa;+ ^fc^a;i*-2Aa;2 + ... + Aa;*- a^i*] A ^alkx^-^^^-~-^x^-''b.x + - + Aa;*-i]Aa?. Z It is clear from this expression that for every value of k the expression (3) has A« as a factor. Moreover the expres- sion (2) for A?/ is the sum of such terms as (3) for different values of k ; and, since each of these terms has the factor Ax', their sum has the factor Aa;. Hence, if we divide Ay by Aa;, we have for the slope Ay/Aa; of PiQ, the expression Aa/ u -f terms with higher powers of Aa;] k=.n. + a,_j[(n - l)a;i"-2 + (^^ - l)(n - 2) ^^,^_3^^ 4- terms with higher powers of Aa?] k = n — \. f a2[2 a^i + Aa;] A: = 2. -h tti A: = L XIX, § 290] GENERAL POLYNOMIAL FUNCTION 451 The slope m of the graph is the limit approached by Ay/Aa; as Aic approaches zero (i.e. as Q approaches Pi along the curve). This gives finally (4) m = ?ia„iCi'*-^ +(n — l)an-i^i''~^ H -f ^ ^2^n + «i [Note that for the values w = 3 and w = 2 this reduces to the expres- sions previously derived for the cubic and quadratic functions. ] Moreover, it follovt^s from the remark above, concerning the fact that Aa; is a factor of A?/, that as Ax approaches zero, Ay approaches zero also. But this proves that f{x) is continuous for every value of x. We have then the theorem : Every polynomial f{x) is a continuous function of x. 290. The Derived Function. In previous cases where we have considered the slope of a curve y =f(x) we have always considered its value at some given point Pi on the curve. As the point Pj moves along the curve, however, the value of the slope in general changes. In other words, the slope itself may be considered as a function of a;. This function is called the derived function or the derivative of f(;x). If the original function is denoted by f{x), the derived function is denoted hj f'{x). In case of the polynomial f{x) considered in the last article the derived function f\x) is obtained from the expres- sion for the slope m by letting the given value Xi become the variable x, i.e. if f(x) = a^x'' + cin-i^'"'^ H h ^i^^ + «05 ^® ^^^® the derived function (5) f\x)= na,x^~^+ (n - l)a„_ia;»-2+ ... + a^. The derived function of any polynomial is readily written down from the following consideration. The derivative of any term a^x'' is Zca^ic*"^ ; i.e, it is obtained by multiplying the term by the exponent ofx and reducing the exponent of x byl. Thus the derivative of a^ is 3 x^, of 10 a;^ is 20 x. The above expres- sion for/' (a;) shows that the derivative of a polynomial is the 452 MATHEMATICAL ANALYSIS [XIX, § 290 sum of the derivatives of its terms. Thus the derivative of 6 a;^ — 3 ic* + 7 a;2 — 1 may be written down at once ; it is equal to 35 a^ — 12 a;3 _j_ 14 ^.^ Observe that the derivative of a con- stant is 0. The relation between the derived function f{x) and the slope of the graph at any point, is expressed as follows : The slope of the graph of the curve y =f{x) at the point x = Xi is equal to the value of the derived function for x=Xi^ i.e. m=f'{x^. Further, since the derived function of a polynomial is a polynomial, it follows from the theorem at the end of the last article, that the derived function of a polynomial f{x) is a con- tinuous function of x. EXERCISES] Find/Cx) when 1. /(a;) = x3 + 4 x2 - 6 a; + 3. 2. /(a;) = 5a:6-4a;8 + 6a;2 + 2a; + l. 8. /(x) = 7 a:7 - 4x3 + 2 X + 19. 4. /(x)=3x6-4x* + 2x8 + 3x2+ 1. 6. Find the equation of the tangent to 2/ = 4 x* ~ 3 x + 1 at (1, 2). 6. Find the equation of the tangent to y = x^ — 5x2 + 2 at the point (1,-2). 291. The Graph of a Polynomial S{x). In drawing the graph of a given polynomial of degree greater than 3, we may proceed as in the cases of polynomials of degrees 2 and 3. There are two general theorems to aid us : (1) The graph of any polynomial is a continuous curve ; in particular, the value of y does not become infinite except when X becomes infinite. (2) The tangent to the graph at any point P turns continu- ously as P moves along the curve ; i.e. the curve has no sharp corners and the tangent is nowhere vertical. (Why ?) We found in discussing the graphs of cubic functions that XIX, § 291] GENERAL POLYNOMIAL FUNCTION 453 the values of x for wMcli the slope is zero were particularly- helpful, ill view of the fact that they gave us, in general, the turning points (maxima and minima) of the graph. Let us apply these principles to an example. Example. Draw the graph ofy=f(x)= -|(3 a^— 4 x^ - 12 «2 + 3). Weliave f(x) = i(x^- x^-2x)= 4x(a;- 2)(x + 1). Hence/(x) = when a; = 0, 2, — 1. We require next a table of correspond- ing values of x and y. Here synthetic division is often convenient. Thus, to find /(ic) when a; = 2, we write 3 -4 -12 3[2 6 4-16-32 3 2-8-161 -29 = 32/. Hence y = — 9| when x : = 2. When a; = 3, we have 3 - 4 - 12 9 15 9 3|3 27 3 6 3 9 30 = 32/. Hence y = 10 when a; = 3. :::::i^-::"t: 22 ^__.5. ==-========;== ::::3::^:::i=::: ;::::i:::|i::i: ^^^ Fig. 251 We may note that since all the partial results 3, 5, 3, 9, 30 are positive, any value of a; > 3 will give values of y greater than 10. Finding the values of y for other values of x, we have the following table: x = -2 -1 1 2 3 y = 111- -1 1 -3^ -H 10 w = We have also indicated in the table the values of x for which m is zero. These data give us the graph exhibited in Fig, 251. This example sug- gests certain other general theorems regarding the graph of a polynomial, which are discussed in the following articles. 454 MATHEMATICAL ANALYSIS PCIX, § 292 292. The Value of a Polynomial for Numerically Large Values of x. In the example of the last article we saw that for all values of x>S, the values oif(x) were greater than 10 ; in fact, the nature of the synthetic division showed that as x increased indefinitely from x = 3, the value of f{x) increased indefinitely. Any polynomial f(x) of degree n with real coefii- cients (§ 288) may be written in the form f{x) = aAl +f^^i=l^' + ^^=?^' + - +-^^1 [_ \ ttna?" anX"" anX'^Jj =a„x.[l+(o„_,l+c„.,l+...+c.i)]. Since the absolute value of a sum is equal to or less than the sum of the absolute values of its terms (§ 35), we have, X'^l 11. < 1 Gn-1- X X'^l 1 Cn — i|(|c,.i| + |c„_2| + - + |eo|)<-^, (kl>l). x\ \x\ where c is a positive number independent of x. Hence, if | a; | >c, the value of the expression in square brackets above is cer- tainly positive. Therefore for sufficiently large values |a;|, the sign oif{x) is the same as the sign of a„ic'». If an is positive and x becomes positively infinite, f(x) is positive. If a„ is positive and x becomes negatively infinite, f{x) is positive if n is even, and negative if n is odd. If a^ is negative and x becomes positively infinite, f(x) is negative. If a„ is negative and x becomes negatively infinite, f(x) is negative if n is even, and positive if n is odd. As X increases indefinitely in absolute value, the value of f(x) increases indefinitely in absolute value. For sufficiently large values of\x\, the sign off{x)is the same as the sign, of anX"". XIX, § 293] GENERAL POLYNOMIAL FUNCTION 455 In particular, this leads us to the following theorems. Iff{x) is a polynomial of even degree, the infinite branches of the graph ofy = f{x) are either both above the x-axis or both below the X-axis (according as a„ is positive or negative). If f(x) is a polynomial of odd degree, the infinite branches of the graph of y =^ f(x) are on opposite sides of the x-axis (below the a;-axis on the left and above the a>axis on the right, if a,^>0 ; above the ic-axis on the left and below on the right, if a^<0). From these theorems and from the continuity of the function f(x) we derive the following corollary. The graph of a polynomial f{x) of odd degree with real coeffi- cients must cross the x-axis at least once and, if it crosses more than once, it must cross it an odd number of times. The graph of a polynomial of even degree with i-eal coefficients either does not cross the x-axis at all or it crosses it an even rmmber of times. 293. The Zeros of a Polynomial /(x). The Roots of the Equation /(x) = 0. A value of x for which f(x) — is called a zero of f{x) ; i.e. if f(b)= 0, then 6 is a zero of f(x). The zeros of f(x) are, therefore, the values of x which satisfy the equation/ (a;) = 0. The zeros of /(a;) are called the roots of the equation f(x) = 0. The factor theorem (§ 261) tells us that if a is a zero of f{x), then a; — a is a factor of f(x). Since a polynomial of degree n cannot have more than n distinct fac- tors of degree one, we may state the following theorem. A polynomial f(x) of degree n cannot have more thann dis- tinct zeros. Since at the turning points of f(x) the slope is always zero, it follows from the fact that the derived function is of degree n — 1 that a polynomial f{x) of degree n cannot have Tnore than n — 1 turning points {maxima andminimxi). 456 MATHEMATICAL ANALYSIS [XIX, § 294 294, The Number of Roots of /(x) = 0. We have seen that every quadratic equation has two roots which may be real or imaginary and which may be equal. We have also seen that every cubic equation f(x) = 0, whose coefficients are real, has at least one real root. If this root be ri, we may write (§ 261), f{x) = (x — rj Qix), where Q(x) is a polynomial of degree 2. The latter has two zeros, real or imaginary, so that any cubic function with real coefficients may be resolved into 3 linear factors, ^(^) ^ ^^(^ _ ^^)(^ _ ,,^)(^ _ ^^y It may be proved that any polynomial {no matter whether the coefficients are real or imaginary) has at least one zero (real or imaginary). This statement is called the fundamental theorem of algebra. We shall accept it as valid without proof, since its proof is too difficult for an elementary course.* From this theorem it is easy to prove the following theorem : Any polynomial f(x) of degree n may be resolved into n linear factors. Proof : By the fundamental theorem, f(x) has one zero. Denote it by r^. The factor theorem then gives f{x)^{x--r,)Q,{x), where Qi is a polynomial of degree n — 1. By the funda- mental theorem, Qi{x) has a zero r2. Hence q,ix) = (oj - ro)Q2(a;), or f{x) = (a; - r;){x - r.,)Q2{x). Again, Q2(x) is a polynomial of degree n — 2. If n > 2, Qg has a zero, say r^, which leads to the expression f{x) = {x- ri) (x - rg) {x - rg) Qz{x), where Qi(x) is a polynomial of degree n — S. Continuing this * This theorem was first proved by Gauss in 1797 (published 1799) when he was 18 years old. For proof see Fine, College Alyelfra, p. 588. XIX, § 294] GENERAL POLYNOMIAL FUNCTION 457 process we find /(a;) = (a; - ri)(a; ^ rj) "• (aJ - r„)Q„, where Qn is a constant which evidently must be a„ if f(x) is a^Xn + ••• + Oq. We have then finally f(x) = a^(x - ri)(x - r.2) - (x - r„). Each of the numbers Vi, r2, •••, r„ is a root of the equation f(x) = 0. This proves the theorem just stated. Moreover, no number different from Vi, rg, •••, r„ can be a root of this equation. For suppose s were such a number, then we should have /(s) = a„(s — ri){s — r2) ••• (s — r„). Since each of these factors is under the hypothesis different from zero, the product /(s) is different from zero. Some of the num- bers ri, r2, •••, r„ may be equal, however. This possibility leads to the following definitions. If f{x) is exactly divisi- ble by a; — r but not by (a; — r)^, then r is called a simple root of f(x) = 0. If f(x) is exactly divisible by (x — ry but not by (a; — r)^ then r is called a dovble root of /(x) — 0. If f{x) is exactly divisible by (a; — r)* but not by (x — r)^^ then r is called a A:-/oM root, or a root 0/ order k. A root of order greater than one is called a multiple root. If /(a;) represents a polynomial, the equation f(x) = is called an algebraic equation. Then we may state the last theorem as follows : Every algebraic equation of degree n has n roots and no more, if each root of order k is counted as k roots.* EXERCISES 1. Is 1 a zero of the polynomial x'' — Sx^ + 2z* — x-\-S? 2. Is 2 a zero of the polynomial a;* — 16 ? 3. Is 3 a root of the equation x^ + Zx'^-{-x-Bz=0? * It is logically necessary to note the fact that, if exactly k of the roots ^i> ^2» — equal r, f{x) is divisible by (k — r)* but not by (a; — r)*+i. Why? 458 MATHEMATICAL ANALYSIS [XIX, § 294 4. Find k so that a; =1 is a root of the equation a;^ + A^x^ — x -f 1 = 0. 6. Pind k so that 2 is a root of the equation x^ + x^ — Ax + 3 = 0. 6. How many roots has the equation x'^ + x8 + x + 3 = 0? How many of these roots are positive ? 7. How many roots has the equation x^— 2 x* + x^ — 3 x2+2 x— 1=0 ? How many of these roots are negative ? 8. Find graphically the real zeros of the functions (a) xs - X. (6) x3 + 2 X - 1. (c) x^ + 3 x + 2. (d) x^ - x2 - 6 x + 8. Draw the graph of each of the following functions : 9. t/ = t^3j [3 x* - 4 x8 - 24x2 + 48 X + 13]. 10. y = i^j [3 x* + 8x8 - 6 x2 - 24 X - 12]. 11. ?/ = ^Jj [3 X* + 4 x8 - 12 x2 + 24]. 12. 2/ = 2 x* - 14 x3 + 29 x2 - 12 X + 3. 13. Prove, without assuming the fundamental theorem of algebra, that every algebraic equation of odd degree with real coefficients has at least one real root. 295. Successive Derivatives. The derived function of a polynomial f(x) of degree n is a polynomial f'{x) of degree n — 1. The derivative of f{x) is a polynomial of degree 71 — 2, is denoted hy f"(x), and is called the second derivative of f{x). Similarly, the derivative of f"{x) is called the third derivative of f(x) and is denoted by f"'{x). Similarly, the fourth, fifth, etc. derivatives may be found. The nth derivative of a polynomial of degree n is evidently a constant. Thus, if /(x)=x*-3x3-7x + 2, we have /'(x) =4x8-9x2 - 7, /"(x) = 12x2- 18x, /'"(x) = 24x- 18, /i^(x) = 24. 296. Taylor's Theorem. The following formula is known as Taylofs theorem: (6) /(x) = /(a) + /'(a)(x - a) +^ (x - af + ... +m(x-ar. n I This formula enables us to express any polynomial in a? as a polynomial in a? — a, where a is any constant. XIX, § 296] GENERAL POLYNOMIAL FUNCTION 459 For example, if we have/(ic)= a:^ _ 4 ^ + 2 and desire to express /(x) in terms of x + 1, we first find /'(x) = 3 x^ - 4 ; /"(x) = 6 x ; /'"(x) = 6. The coefficients in the above formula are, for a = — 1, /(-l)=5, /'(-1) = -1, /"(-l) = -6, /'"(-1) = 6. Therefore we have, from (6), x3 _ 3x + 4 = 5 -(X + 1)- 3(x + l)H(x + 1)3. Proof. We have seen in § 290 that the derivative of a;* is kx^~^. Likewise the derivative of {x — a)* is k(x — ay~\ For if y z=i{x — ay, we have, as in § 289, y + Ay ={x-\- Ax— a)*=[(a;— a)+ Ace]* = (x — a)*+ k(x — ay~'^Ax + terms with a factor Aaj^. Hence -^ = k{x — a)^^ + terms with a factor Ax, Ax and the limit of Ay /Ax is obviously k(x — tt)*~^ Let us now set (7) /(aj)=4, + A(«- a) 4-^2(0; -a)2+ - + J.,(a;- a)* +-. We then have, by taking successive derivatives of both sides, f(x) = ^li + 2 A2{x - a) -f ... + kA,{x - ay-' + ..., r(x) = 2 ^2+ - + A;(A^ - l)A,(x - af-' + ..-, /(*)(a;) = Zc ! ^;t + terms containing {x — a) as a factor. These relations must all be true for all values of x ; hence they must hold when x = a. But this gives f{a)=A„ f'{a) = Au /»=2^2,-, /*(«) = A: ! ^„ -. Hence By substituting these values in (7) above, we obtain Taylor's 460 MATHEMATICAL ANALYSIS [XIX, § 296 theorem as given in relation (6). Another form of Taylor's theorem is obtained by replacing a; by a; + a in relation (6). This gives (8) /(x + a) = f(a) +/'(fl)x +^-^x^ + ... +i^x^. EXERCISES 1. Write down the successive derivatives of the following polynomials : (a) x8 + 4a;2_i2a;+l7. (6) 2x*-3a:8 + 8x2-14x + 18. (c) a;5 + 2 a;- 1. (d) 1 -3x +4x2 + 5x8. 2. Prove that the nth derivative of a^x" + an-ix"-i + ••• + aix + oo is equal to OnTi I . 3. Expand each of the following by Taylor's theorem : (a) x3 + 4 x2 — 12 X + 17 in terms of x - 1. (6) 2 X* - 3 x3 + 8 x2 - 14 X + 8 in terms of x - 2. (c) x^ + 2 X — 1 in terms of x + 1 . Id) 1 — 3 X + 4 x2 + 5 x3 in terms of x + 2. 4. By relation (8) in § 296 express each of the following as a polyno- mial in X : (a) /(x-l)if/(x)=x5 + 4x2-12x + 17. (h) /(x-2) if/(x)=2x4-3x8 + 8x2-14x + 8. (c) /(x + 1) if/(x)=x6 + 2x-l. W /(x + 2)if/(x) = l-3x + 4x2 + 6x8. 297. Multiple Roots. If we apply Taylor's theorem succes- sively to/(ic) and/'(ic), we obtain (9)/(») = (10) /'(») = f\a) +f"{a){x - a)+i^ {x - ay+^^ix - ay+ ..... XIX, § 297] GENERAL POLYNOMIAL FUNCTION 461 If /(a)= 0, the first relation shows that x—a is a factor of /(a;) ; this constitutes a new proof of the factor theorem. If /(a;) 13 divisible by a; — a but not by {x — ay, it follows that /(a) =n d and that f'{a) ^ 0. Hence by (10), or by the factor theor3n_, fix) is not divisible by a; — a. If f{x) is divisible by (x — a)- but not by (x — df, we have, from (9), /(a)=0, /'(a)=0, f"{a)=^0. We then conclude from (10) that if a is a double root of f{x) — 0, it is a simple root of f'{x) = 0. In general, if f(x) is divisible b}^ (x — a)* but not by (x— a)*+\ relation (9) shows that /(a) =/'(«) =/''(«) =- =/'-'(«) =0; /^a) =?t 0. Hence, by (10), f'(x) is divisible by {x — a)*~^ but not by (a? — a) *. This leads to the following theorem. A simple root of f(x)= is not a root of f\x)=0. A double root of f(x)= is a simple root of f'{x)= 0. In geyieral, a root of order k off(x)= is a root of order k — 1 off'(x)= 0. The following corollary of this theorem is evidently true. Any multiple root off(x)==0 is also a root off\x)=0. If f(x) andf'(x) have no common factor, f(x) = has no multiple roots. If(f> {x) is the H. C. F. off(x) and f'{x), the roots of <ft (x)=0 are the multiple roots off(x) = 0. Example 1. Examine for multiple roots the equation /(x) = x3 4- x2 - 10 x + 8 = 0. We have /' {x) =Sx'^ + 2x- 10. To find the H. C. F. of /(x) and f\x) we proceed as in § 259 : 8a8-f3a:2-30x+24 3x8 + 2x2- lOx x + 1 3x X2- 20 X + 24 8x2- 60 X + 72 8x2 + 2x -10 -62X + 82 3x2 + 2x- 10 186x2 + 124x -620 -186x2 + 246X 370 X -620 It is now clear that /(x) and /'(x) have no common factor, we conclude that/(x)= has no multiple roots. Hence 462 MATHEMATICAL ANALYSIS [XIX, § 297 Example 2. Examine for multiple roots the equation /(x)=x*-2x3 + 2x-l = 0. We have /'(x) = 4 x^ - 6 x^ + 2. 2x*- 2x*- -4x8 + 4x -3x8+ a; -2 - x8 + 3x 2x8-6x 2x8-3x2 -2 + 4 + 1 3x2-6x x2-2x + 3 + 1 X 2x3-3x2 + 1 2x3-4x2 + 2x x2 _ 2 X + 1 a;2 _ 2 X + 1 2x4-1 Hence (x — 1)2 is the H. C. F. of /(x) and /'(x), i.e. x = 1 is a triple root of / (x) = 0. The fourth root of /(x) = is x = — 1. How is it obtained ? EXERCISES 1. Examine for multiple roots each of the following equations : (a) x3-3x2-24x-28=0. (6) x* + xS + 1 = 0. (c) x5-7x3-2x2 + 12x + 8 = 0. Id) x5 + X* - 9 x8 - 5 x2 + 16 X + 12 = 0. (e) x4-6x8 + 12x2-10x + 3 = 0. (/) x8-3x6 + 6x8-3x2-3x + 2 = 0. 2. Prove that the graph of y =/(x) is tangent to the x-axis at a point representing a multiple root. 3. Prove that the graph of 2/=/(x) crosses or does not cross the X-axis at a point representing a multiple root according as the order of the root is odd or even. [Hikt : Use Taylor's theorem.] 4. Prove that a root of order k of /(x) = is a simple root of /*-i(x)=0. 298. Complex Roots. If a -\- hi (a, 6 real numbers, i2= — 1) is a root of an algebraic equation /(«) = with real coefficients, then a — bi is also a root of the same equation, By hypothesis a -f- 6i is a root of the equation f{x) = anx- + a„_iaJ"-^+ ••• + ao = 0, i.e. f{a + 60 = a^ia + biY+ a„_i(a + bif-^^ + ••• -f Oo = 0. If each of the terms in the preceding expression be expanded XIX, § 298] GENERAL POLYNOMIAL FUNCTION 463 by the binomial theorem, the powers of i reduced to their lowest terms (i^ = — 1, ^s = — i, etc.), and terms collected, we obtain ^, , t-\ n , r\' where P represents the sum of the terms independent of i ajid Q is the coefficient of L But since P + Qi = hj hypothesis, it follows from § 281, that both P = and Q = 0. We wish to prove that a — bi is a root of f(x) = ; i.e. f{a — bi) = 0. To prove this we merely have to notice that f(a — bi) may be obtained from the expression for /(a 4- bi) by replacing i by — i. Therefore f{a-bi)=P- Qi, where P and Q represent the same quantities as before. But we have just shown that P = and Q = 0. Therefore /(a — bi) = or a — 6i is a root oif(x) = 0. EXERCISES 1. Solve X* + 4 a;3 + 5 a:2 + 2 X - 2 = 0, one root being - 1 + i. 2. Solve a!:* + 4a:S + 6a;2-f4x + 5=:0, one root being i. 3. Solve a;4 - 2 x3 + 5 a;2 — 2 a: + 4 = 0, one root being 1 — i Vs". 4. If a+ Vb (a and b rational but V6 irrational) is a root off(x) = with rational coefficients, a — V6 is also a root. [Hint : Show that /(a + y/b) reduces to the form P + Qy/b where P and Q contain only integral powers of b and Q is the coefficient of Vb. Since P + QVb = 0, P = and Q = 0. Why ?] 5. Solve 2 x* - 3 a;3 - 16 x2 - 3 X -f 2 = 0, one root being 2 + VS. 6. Form an equation with rational coefficients, of which two of the roots are i and 1 + \/2. 7. Solve the equation x^ - (4 + V3)x'^ + (5 + 4 V3)x - 5V3 = 0, if one root is 2 — i. 8. Solve the equation x^ — (5 + i) x^ + (9 + 4i)x — 5 — 5 1 = if one root is 1 + i. Is 1 -- I a root ? 464 MATHEMATICAL ANALYSIS [XIX, § 299 299. To Multiply the Roots of an Equation. To trans- form a given equation f(x)= into another whose roots are those of /(a;) = each multiplied by some constant 7c, multiply the second term of f{x) by Jc, tJie third term by k^, and so on, taking account of the missing terms if there are any. The required, equation is f(y/k) = 0. For, if f(x) vanishes when X = a,f(y/k) will vanish when y = ka. Hence, if the given equation is a„a;" -f a^_^x''~^ + ... -f ao = 0, the required equation is ""(!)" +""-■©"""'■■■ +""='' which on multiplication by k"" becomes ttn?/'* H- kan-if"-^ + k'^an-2y''-^ + • • • + k^ao = 0. If A: = — 1, we have, the roots of f(^x) = are equal respec- tively to those off{x) = with their signs changed. Example 1. Transform x^— ^x"^ -{- 5 = into an equation whose roots are twice those of the given equation. The desired equation is x^ - 4(2)a;2 + 6(2)3 = 0, or x^ - 8 x2 + 40 = 0. Example 2. Transform x'' — Sx^-\-4iX^-2x-\-l = into an equa- tion whose roots are those of this equation with their signs changed. The result is (-a:)7-3(-x)5 + 4(-x)4-2(-x) + 1 = 0, or a;7 _ 3 x5- 4 x* - 2 x - 1 = 0. EXERCISES Obtain equations whose roots are equal to the roots of the following equations multiplied by the numbers opposite. 1. x6 - 2 x^ + x + 1 = 0. (2) 3. x4 - x2 + X + 1 = 0. (-3) 2. x7-6x8+2x- 1 = 0. (-2) 4. x^ + x* -x^ + x- 1 = 0. (2) Obtain equations whose roots are equal to the roots of the following equations with their signs changed. 5. x7 - 6 x« + 2 X* - X + 1 = 0. 7. x7 - x« + x6 - X* - 2 = 0. 6. xi6 _ 1 = 0. 8. 1 - X - xa - x8 - X* - x6 = 0. XIX, § 300] GENERAL POLYNOMIAL FUNCTION 465 300. Variations in Sign. A variation of sign or change of sign is said to occur in f(x) whenever a term follows one oi opposite sign. Thus the equation o^ — 3 a;^ 4-7=0 has two variations of sign. If f(x) has real coefficients and is exactly divisible by x — 7c, where k is positive, then the number of variations of sign in the quotient Q{x) is at least one less than the number of variations of sign infix). Before proving this statement let us consider the process of dividing /(a;) = x^-{- x^ — 3 x^— 2 x^ — x^ + b x—lhj x — 1 and /(x) = a;^ — a^-f- 4 a;2— 13 cc -h 2 by a; — 2, making use of synthetic division. • 1 1 _3 _2 -1 5 -1 g. 1 2-1-3-4 1 Q(a;) = l 2-1-3-4 1 1 _1 4 -13 2 |2 2 2 12 -2 Q(a!)=l 16-1 It will be noted in these examples that Q(x) has no varia- tions except such as occur in the corresponding or earlier terms of f{x) and that since f(x) is exactly divisible by the given divisor, the sign of the last term of Q{x) is opposite to that in f{x). Let us now prove the statement in general. Proof : From the nature of synthetic division it follows that the coefficients in Q(x) must be positive at least until the first negative coefficient of f{x) is reached. Then, or perhaps not until later, does a coefficient of Q{x) become negative or zero, and then they continue negative at least until a positive coefficient in f{x) is reached. Therefore Q{x) has no variations except such as occur in the corresponding or earlier terms of f(x). But by hypothesis f(x) is exactly divisible by a;— A; and 2h 466 MATHEMATICAL ANALYSIS [XIX, § 300 hence the sign of the last term in Q(x) must be opposite to that in f(x). Therefore the number of variations of sign in Q(x) must be at least one less than the number of variations of sign in /(a;). 301. Descartes's Rule of Signs. Theequationf{x)=0 with real coefficients can have no more positive roots than there are variations of sign in f(x) and can have no more negative roots than there are variations of sign inf(— x). Proof: Let r^, ^2, — , ^^(i? ^ ^) denote the positive roots of f(x) — 0. If we divide f{x) by a; — ri, the quotient by ic — rj, and so on until the final quotient Q{x) is obtained, then we know from the last theorem that Q(x) contains at least p fewer variations of sign than f{x). But the least number of varia- tions of sign that Q{x) can have is zero. Therefore f{x) must have at least p variations, i.e. at least as many variations as f{x) =0 has positive roots. Second, by § 299, we know that the negative roots of /(a;)=0 are the positive roots of /(— a;) = and, hence, by the first part of this proof, we know that their number cannot exceed the number of variations of sign in/(— «). It is important to notice that Descartes's rule of signs does not tell us how many positive and how many negative roots an equation has. It merely tells us that an equation cannot have more than a certain number of positive roots, and cannot have more than a certain number of nega- tive roots. Example. What conclusions regarding the roots of the equation /gT _ 4 x5 + 3 a;2 — 2 = can be drawn from Descartes's rule ? The signs of /(a;) are -f 1 — , i.e. there are 3 variations and hence the equation has no more than 3 positive roots. The signs of /( — ic) are h + — , i.e. there are two variations and hence the equation has no more than 2 negative roots. But the equation is of degree 7 and has 7 roots. Therefore the equa- tion has at least two imaginary roots. Can there be more than two imaginary roots ? XIX, § 303J GENERAL POLYNOMIAL FUNCTION 467 EXERCISES What conclusions regarding the roots of the following equations can be drawn from Descartes' s rule ? 1. x^ - 2 x6 + X* - 1 = 0. 4. x^ - 2 X* + x^ - x^ - X + 1=0. 2. x^ + X* — x^ + 1 = 0. 5. x** - 1 = 0. (w odd) 3. x23 - 34 xi2 + X — 45 = 0. 6. x** - 1 = 0. (n even) 7. Show that the equation x^ — 5x2 — x + 10 = has at least two imaginary roots. How many may it have ? 8. Show that the equation x^ + x^-i-x — 1=0 has two and only two imaginary roots. 9. Show that the equation x^ + 4 x' + 2 x — 10 = has six and only six imaginary roots. 10. Can you tell the nature of the roots of the equation x* + ix^ — 3 ix + 4 = 0? 302. Equations in ^-form. If each term of the equation f{x) = a„a;" + a,,_,x^-' + ... + a^ = is divided by a„ (by hypothesis an ^ 0), we obtain the equation x^ + p^x''-'^ + PiX""-^ + ... + jp^ = 0, in which the leading coefiicient is unity and p^ = -^^, etc. An equation in this form is said to be in the p-form. For many purposes this is the most convenient form. 303. Rational Roots. A rational root (^0) of the equation f{x) = when the equation is in the p-form with integral coeffi- cients is an integer and an exact divisor of the constant term. Proof. Suppose that the equation f(x)= has a root a/b where a/b {b > 1) is a rational fraction in its lowest terms. Then we have (11) (tf + p/tT' + ■■■+ pUt) +P, = 0. I, - " \b 468 MATHEMATICAL ANALYSIS [XIX, § 303 Multiplying both, members of (11) by 6"~^ we have or (12) T^~ (-^1^""' + ihci^'-^h + - + Pnh^"), The right-hand member of (12) consists of terms each of which is an integer. The left-hand member of (12) is a fraction in its lowest terms. Therefore the assumption that the fraction a/h is a root of f{x) = leads to an absurdity. Now suppose r (:^ 0) is an integral root. Then rn -hpir"-! -f p^r^-^ -+- ••• -h i?« = 0. If we transpose the constant term pn and divide by r, we obtain (13) r^-i^p^r^-2+ ... -|.p^_^ = _-P2. r Now each term of the left-hand member of (13) is an integer ; hence pn/r must be an integer, i.e. pn must be exactly divisible by r. 304. To Find the Rational Roots of an Equation with Ra- tional CoeiKcients. If the equation is not in the p-form with integral coefficients, reduce it to that form and then make use of the results in § 303. The following examples will explain the methods. Example 1. Solve the equation x^ + S x^ — 4 x ^ 12 = 0. By Descartes's rule of signs we know that the equation has no more than one positive root and no more than two negative roots. From the last article we know that if the equation has rational roots they are factors of 12. Thus we need only try 1, — l, 2, — 2, 3, - 3, 4, — 4, 6, - 6, 12, -12. XIX, § 304] GENERAL POLYNOMIAL FUNCTION 469 By synthetic division we have 1 3-4-12 12 2 10 12 I 16 6 The depressed equation * ia x^ -\- ^ x + 6 = (x -^ S)(x + 2)= 0. There- fore the roots of the original equation are 2, — 3, — 2. Example 2. Solve the equation 2a:8_^a;2 + 2x + l = 0. Writing the equation in the |)-f orm we have X3 + I X2 + X + ^ = 0. If we multiply the roots of this equation by k, we obtain x^-\--kx^-\-k'^x + — =0. ^ z If we choose k equal to 2, this equation becomes (14) x^-\-x^-h4:X + 4 = 0, an equation whose roots are twice those of the original equation. By Descartes's rule of signs equation (14) has no positive roots. Any rational roots are then negative, and are factors of 4, i.e. — 1, — 2, — 4. By synthetic division 1 14 4 |-1 -10-4 10 4 The depressed equation is a;^ + 4 _ q. Therefore the roots of (14) are — 1, 2 1, — 2 I and the roots of the given equation are — ^, i, — i. EXERCISES Solve each of the following equations. 1. ic« + 5 x2 -f- 15 X + 18 = 0. 4. 6 a;3 + 7 x2 _ 9 x + 2 = 0. 2. xs + x2 + X + 1 = 0. 5. 6 x3 - 2 ic2 + 3 X - 1 = 0. 3. x=i + x2-4x-4 = 0. 6. 2x4 + 3x3-10x2-12x+8=0. Find the rational roots of each of the following equations. 7. X* - 3 x2 - 4 = 0. 10. 2 X* - x8 - 5 x2 + 7 X - 6 = 0. 8. x6 - 32 = 0. 11. 2 X* + 2 x3 - x2 + 1 = 0. 9. X* + x8 + x2 + X + 1 = 0. 12. 4 x* - 23 x2 - 15 X + 9 = 0. * If r is a root of a given equation /(.x*) = and f(x) = (x — r)Q(x), then the equation Q{x)= is called the depressed equation. 470 MATHEMATICAL ANALYSIS [XIX, § 305 306. The Solution of an Equation with Numerical Coeffi- cients. The preceding articles furnish a number of methods for attacking the problem of finding the roots of an algebraic equation/ (a;) = with given numerical coefficients. (1) We may examine the equation for multiple roots (§ 297). (2) If the equation f{x) = has rational coefficients, we can find all the rational roots by a finite number of trials. (3) When any root a has been found, we may divide f(x) by X — a and thus make the finding of the remaining roots depend on an equation of lower degree (the depressed equation). 306. Irrational Roots. Graphical Approximation. In order to compute approximately any one of the real irrational roots of an equation/ (a;) = whose coefficients are real numbers, we require first a rough approximation to the root which is to be computed. The graph of y =f(x) is a powerful tool for this purpose. An example will make the method clear. Example. Locate approximately the real roots of the equation f(x) = x6 - ISx'-^ + 2 a; + 5 = 0. A table of corresponding values of x and f(z) is as follows. I^' -4^ 7ko_ ::::^:i::i-:: :i:::|?:i::d=: ::::4?:i::: i: -^ f- 1)^^ uy :E/:E:::E:E:E: X -2 - 1 1 2 3 /(a;) -83-11 6 -5 - 11 137 Fig. 262 Figure 252 exhibits a rough graph of this function constructed from this table. We conclude that a root of the equation lies between — 1 and 0, another between and 1 , and a third between 2 and 3. Moreover Descartes 's rule tells us that this equation can have no more than two positive roots and no more than one negative root, since there are only two changes of sign in f(x) and only one in/(— x). We have therefore located all the real roots of this equation. XIX, § 307] GENERAL POLYNOMIAL FUNCTION 471 A more accurate construction in the neighborhood of one of these points enables us to get a better approximation. For example, the values x = 2.2 and 2.3 give us respectively y = — 1.97 and 6.21. By drawing a smooth ~ -T-- 4* -. ^^ " ^: : ^^ -rr ,t 2-0 -2- Ir- 2-e^' ,../ ' ^ i* ^^' ,'' ** , * *' — *^^ - - __ - "- - -■ Fig. 253 curve through the three points corresponding to x = 2, 2.2, 2.3 (plotted on a large scale, Fig. 253) we may estimate the root of f{%) = to be ap- proximately 2.23. 307. Newton's Method of Approximation. Having found a first approximation to a root of an equation /(«)= 0, we may secure a better approximation by a method first suggested by Sir Isaac Newton (1642-1727). In Fig. 254 let GC represent the graph of y—f{x) in the neighborhood of a root a? = a of the equation. Let OM^ — x^ represent the approximation to the root found ; let M^Px = ^1= /(xi). Let the tangent to the graph at Pi(iCi, y-^ cut the a;-axis in T. The abscissa OT will then, in general, be a much closer approximation to the desired root. The equation of the tangent at Pj is (15) 2^-/(^i) = /'(^0(^-a^i). Placing 2/ = and solving for x we have Fig, 254 (16) 052= OT=Xy /fe) 472 MATHEMATICAL ANALYSIS [XIX, § 307 where x^ denotes our second approximation. We have then (17) ^2 = JCi + ^1, where the correction hi is given by (18) K = -f^. Example. Find by Newton's method a better approximation to the root X = 2.23 of the equation x^ — 13 x''^ + 2 ic4- 5 = discussed in § 306, f(x) = ic5-13x2 + 2a; + 5. /'(x) = 5a:*-26x + 2. /(xi)=/(2.23)=-0.039.» /'(xi)=/'(2.23)=67.67.» Hence we have ^~ /'(a^i) 67.67 whence X2 = 2.23057. 308. The Accuracy of Newton's Method. A question that nat- urally arises is : How accurate is this root, i.e. to how many decimal places is it correct ? Taylor's theorem gives us information on this point. We have ,^ (19) /(xi + ;iO = /W + /'(^i)'^i+'^-^^^i'+ -. If our first approximation to the root is x = Xi and hi is the correction, f Newton's method gives to ^i a value which makes the sum of the first two terms of Taylor's expression vanish. Since h\ is very small, the terms beyond the third (involving h\^ and higher powers of ^i) are insignificant fii (xA compared with the term , . h^. Hence for our purpose we may write 2i I (20) /(a:i + Ai) = i/"(a;i)A;2. In the example considered above we have /"(x)= 20x3-26. /"(a^i) =/"(2.23)= 195.8. hi^ = (0.00057)2 = 0.00000032. Hence we have \f"{x{)hi^ =/(xi + h{) = 0.0000313. ♦ Use synthetic division to get these values. t In the example just cousidered ^i = 0.00057. XIX, § 309] GENERAL POLYNOMIAL FUNCTION 473 Moreover ^,^^^ ^ ^^^ ^^,^^^^ +r{x{)h, + ..-, and in this example /'(aii + hi) = 67.67 + 0.11 = 67.78 approximately. It follows that the new correction is about ;i2 = ^/(?l+M<_ 0.000001. f'{xi + hi)^ Therefore we may conclude that x = 2.23057 is the root sought, to five decimal places. EXERCISES Find to three places of decimals the irrational roots of the following equations. 1. a;3 + 3x + 20=0. 2. a;^ + 2x2 - a; + 3 = 0. 3. a^ + a;-l = 0. 4. a:8 + 4 a;2 _ 6 = 0. 5. x^ + Sx"^ - Sx-1 = 0. 6. If X is the cosine of an angle and y is the cosine of one third of the angle, then 4 y^ = 3 ?/ + x. Find the value of cosine of 20° to three places of decimals. 7. An open box is to be made from a rectangular piece of tin 9 x 10 inches, by cutting out equal squares from the corners and turning up the sides. How large should these squares be so that the box shall contain 59 cu. in,? 8. Find the cube root of 12 ; 45 ; - 37. 309. The Relation between Roots and Coefficients. If fit ^2) •••? ^„ are the roots of the equation x"" + picc""^ 4- P2^"~2 _|_ — hPn = 0, then x^ -{- PiX""-^ -\- PiX""-^ + ... -\-p„=(x-r{){x-r2)-{x-r,;). If we carry out the indicated multiplication in the right-hand member and equate the coefficients of like powers of x, we ' have (21) Pi = -ri-r2 r,, (22) JP2 = nrz + nra + ... + nr^^ + r^n + - + »*n-iV (23) Ps = - ^1^2?*3 - r^r^U ^n-2^n-l^n. (24) P«=(-l)Vir,...r,. 474 MATHEMATICAL ANALYSIS [XIX, § 309 That is, — Pi= the sum of the roots. JP2 = the sum of the products of the roots taken two at a time. — Pg = the sum of the products of the roots taken three at a time. (— lyPn = the product of all the roots. We have at once the following corollaries : 1. To transform an equation into another whose roots are those of the original equation each midtiplied by m, multiply p\ by m, p2 by m^, p^ by m^, and so on (§ 299). 2. To transform an equation into another whose roots are equal to those of the original equation with their signs changed, change the signs of the alternate terms, beginning with the second. Example 1. Solve the equation 2y:^— x'^ — 9>x + ^ = Q given that two of the roots are equal in absolute value but opposite in sign. Let the roots be r, — r, and s. Then r — r + s = \, rs — rs — r^ = — 4:, -r2s=-2. Therefore s = ^ and r = 2 or — 2, i.e. the roots are ^, 2, — 2. EXERCISES 1. Solve a* + x* — 4 5c — 4 = 0, given that the sum of two of the roots is zero. 2. Solve x^ — 6 a;3 — 9 x2 + 54 x = 0, given that the roots are in arith- metic progression. 3. Solve X*- 16 x3+ 86 x^- 176 x + 105 = 0, given that the sum of two roots is 4. Ans. 1, 3, 5, 7. 4. Solve 4 x8 — 20 x2 — 23 X — 6 = 0, two of the roots being equal. 5. If n, ro, rs are the roots of x^ — 5 «'-* + 4 x — 3 = 0, find the value of each of the following expressions : (a) ri2 + rz^ + n^. (6) ri^ -\- r2^ + ra,^ (c) riV + ri^rs^ + r^^rs^. (d) ri^ri + r-i^r^ + r2'Vi + rz'^rs + rs^i + r^^r^. CHAPTER XX DETERMINANTS 310. Determinants of the Second Order. Expressions of the form aib2 — a^bi , where a^, tt2, 61 , 62 ^'^^ ^^J numbers, arise often in mathematical analysis. Thus the area of a triangle with one vertex at the origin and the other two vertices at the points (ai , bi), (a2 , 62)? is equal to ^{aib^ — a26i) (§ 195). Again, the solution of a pair of simultaneous linear equations in two unknowns (§ 69) can be written as two fractions whose numer- ators and denominators are all of this form. (§ 311.) The expression aib^ — ^2^1 ^^J ^^ written in the form I ai bA and is then called a determinant of the second order. Such a determinant contains two rows and two columns. The numbers «i > «2 J ^1 J ^2 J are called the elements of the determihant. The two elements ai, 62 form the so-called principal diagonal. To evaluate a determinant of the second order, i.e. to find what number it represents, one merely has to subtract from the product of the terms in the principal diagonal the product of the other two terms. Thus we may write tti bi 0"2. ^2 = aib^ — a^bi ; 4 7 3 -6 = (4)(-6)-(3)(7)=-45. It is important to notice that each term of the expansion contains one and only one element from each row and one and only one element from each column. 475 476 MATHEMATICAL ANALYSIS [XX, § 311 311. Simultaneous Equations in Two Unknowns. Let the equations be (2) 1 OgflJ + 62S/ = C2. If we solve these equations by the usual method of elimination (§ 69), we obtain (3) C162 ~ ^2^1 ^l^ — ^2^ "" aib2 — a^hi ' "" 0162 — a2&i ' provided a^h^—a^hi^O. We at once recognize the fact that these results may be written in the form (4) x = Ci 61 ai Ci C2 62 02 C2 «! h '} y — «! &1 ^2 &2 02 &2 provided 0162 — c^h =^ 0. The following points should be noted in the above solution. (1) The determinants in the denominators are identical and are formed from the coefficients of x and y in the original equations. . (2) Each determinant in the numerator is formed from the determinant in the denominator by replacing by the constant terms the coefficients of the unknown whose value is sought. Example. Solve by determinants the simultaneous equations (2x-y = l, [Sz + 2y = S. Solution : 1 -1 2 1 3 2 6 = 7' y=- 3 3 2 -1 2 -1 8 2 3 2 XX, § 311] DETERMINANTS 477 EXERCISES Evaluate each of the following determinants ^ 14 61 ^ I — sin a — cos a I 1 • 3. 1 3 1 1 I cos a sin a I 12 a 61 Itan^ sec^l c — d\ ' sec ^ tan d I sin e cos 6 sin a cos a 2. 6. Show that the normal form of the equation of a straight line (§ 205), may be written in the form .* " Up. — sm a cos a \ Solve by the use of determinants the following pairs of equations : f2x + y = 3, g \ 6x^ y = 4. Ax- Sy_ = 2, ¥— • 9. x + y = l, 3 *^ 3 10. 11. ic sin ^ 4- y cos ^ = sin 0^ xcosd ■{■yemd = cos 6. X -\- y tan ^ = sec^ 6, X sec2 d +yctne = sec* ^ + 1- ^ns. 1, tan e. Prove the following identities and state in words what they show. 12. 13. 14. 15. 16. ai &i Ol 02 0,2 62 61 62 dl Oi 02 O2 = 0. ai 61 a2 62 mai ?)i maa 62 (ai + 5i) 61 (a2 + &2) &2 = r» 61 ai 62 02 «i &i 02 62 aa 62 * For example, Ex. 12 shows that in a second-order determinant if the cor- responding rows and columns are interchanged, the value of the determinant is not changed. 478 MATHEMATICAL ANALYSIS [XX, § 312 312. Determinants of the Third Order. To the square array bi Ci (5) O2 C2 ^3 Cs we assign the value (6) aidgCg + azhci + ag^iCz — 016302 — aa^iCg — a^bzCi and the name determinant of the third order. The expression (6) is known as the development or expan- sion of the determinant, the numbers ai, hi, etc., as the elements, and the elements Oi, 62, Cg as the principal diagonal. It is important to notice that in the development (6) each term consists of the product of three elements, one and only- one from each row and one and only one from each column. An easy way of obtaining the expansion (6) of the deter- minant (5) is as follows : Form the product of each element of the first column by the second-order determinant formed by suppressing both the row and column to which the element belongs. Change the sign of the product which contains the element in the first column and the second row and take the algebraic sum of the three products. Example 1. Example 2. Example 3. a\ 61 ci 62 Co hx Ci &1 Ci ai 62 C2 = a\ -a'i + as hz ca hz cs 62 C2 az 63 C3 ai&2C3 — ai&3C2 — «2^iC3 + «2&3Ci + a^biCi — azbiCi. 2 3 -6 4 4 -1 3 1 -1 6 + 6 il + ^U = 2(4 + 7)-H 6(3 + 2) + 4(21 - 8)= 99. = 3 = 3(_2-20): m. XX, § 312J DETERMINANTS 479 EXERCISES Evaluate each of the following determinants. 1. 2 1 3 1 1 -1 1 1 2 2. -1 1 5 -2 -2 2 . 3. 6 -3 3 5 1 7 -4 41 la 6 c 6 <r c c a 6 6. In § 196 it was shown that the area of the triangle whose vertices are Pi(a:i, yi), P2(2C2, 2/2)1 ^3(^:3, tjs) is ^[Xi?/2 - X2?/l + 3^22/3 - X3?/2 + Xa^l - Xi^s]- Prove that the area of this triangle is xi yi 1 I X2 y2 I ■ X3 VZ 1 6. Using the result of Ex. 5, find the area of the triangle whose vertices are (a) (2, 1), (3, !),(-!, 7); (6) (3,2), (3, 6), (-1,-4); (c) (0, a),.(0, -a), (&, 0). 7. Prove that the three points Pi(xi, ?/i), P2(a;2, 2/2), P3(a53» ys) are coUinear if, and only if. a^i yi 1 X2 y2 1 X3 yz 1 8. By means of determinants show that the three points (a, & + c), (&, c -\- a), (c^ a + b) are collinear. 9. By use of determinants determine whether the three points (0, 0), (1, 1), (5, 6) are collinear. 10. Prove that the equation of the straight line through the points Pi(a:i, t/i), P2(X2, 2/2), is = 0. X y 1 Xi yi 1 X2 2/2 1 11. By determinants find the equation of the straight line through each of the following pairs of points! (a) (2, 1), (3, 7) ; (6) (6, 1), (2, - 1); (c) (7, 1), (9, 1). 12. Find by the use of determinants whether the three lines 3 a;— y — 7 = 0, 2x4-j/ + 2=:0, a: — y = are concurrent or not. 480 MATHEMATICAL ANALYSIS [XX, § 313 equations be (7) 313. Solution of Three Simultaneous Equations. Let tha aiOJ 4- 6i2/ + CiZ = dj, If we solve these three simultaneous equations by the usual method of elimination, we obtain, dibzC^ + dzbsCi + dsbiCz — dzb2Ci — dih^^Cz — d^bic-^ (8) ~~ ciiboCs + ota^gCi + a^biCz — a^bzCi — aib^Cz — azbic^ _ aid2Cs + a2C?3Ci + a^diC2 — a^doC} — ai(?3C2 — aodiC^ z = «]&2C?3 + «2^3C?1 + «3^1<^2 — «3^2f'l «i&2C3 + (hbsCi -\- (136102 — 0362^1 O163CZ2 — OL^id^ ^1 &1 Ol C?2 62 Ci C^3 &3 C3 «1 bi Cl tti b. C2 <h h C3 cti ^1 Cl <H (i2 C2 V —r. as d. C3 if «! bi Cl a2 &2 C2 ^3 ^^3 C3 ai &i C?i ag &2 d, ^3 63 C?3 ai ^^1 Cl ag 60 C2 03 &3 C3 provided the denominator of each fraction is not zero. These results may be written in the form (9) Each denominator is the same determinant, which is called the determinant of the system. It is made up of the coefficients of X, tfj z. Each determinant in the numerator is formed from the determinant in the denominator by replacing the coefficients of the unknown whose value is sought by the constant terms. Compare this rule with that given in § 311. ExAMPLB. Solve the following equations by determinants : f6x-22=-2, -3y-4;2 = 7, 2x-6j/ = -10. XX, § 313] DETERMINANTS 481 Solution. -2 -21 7 -3 -4 -19-5 ol u -2 -3 -4 -5 224 112 = -2;2/ = 5 -2 -2 7 -4 2 -19 -336 5 -2 - 112 -3 -4 2 -6 5 -2 -3 7 2 -5 -19 5 -2 -3 -4 2 — 5 448 -112 3. EXERCISES Expand each of the following determinants : 4. Solve by determinants each of the following sets of equations ; 2 3 4 -1 -1 4 3 2 1 1 1 1 a h c a2 62 c2 -7 1 2 -2 4 2 2 ► -6 4 a h g h b f g f c a; z z a; y y y z X [4aj4-5y + 22r = 20, 6. j 3 X - 3 ?/ + 5 5! = 12, [5x + 22/-40=-3. Ans. (1, 2, 3). [x-^y ■\-z=\, 8. ax -f- 6y + cs = d, [ cC^x + 6'^?/ 4- c-z = d^. 10. Solve the equation 7. 3x + y-z = 3, x + y + z =7, 2x + 4y -\-z=12. ax + y — z = a'^ + a — l, — x + ay-\-z = a^— a-\-l, x — y + az = a. Ans. (a, a, 1), 1 -1 5 0. 11. Solve for x and y the simultaneous equations 2i x+1 3 y 2 1 X -1 2 1 = 0, X -2 y 0. 482 MATHEMATICAL ANALYSIS [XX, § 313 12. Evaluate the determinant sin a cos /S 1 cos a sin /3 1 1 1 1 Prove the following identities and express in words what they prove. See Ex. 12-16, pp. 477. 13. 15. 17. «! &i Ci a^ hi C2 az h C3 «! tti &i Gi a^ bi as «3 &3 ai tti «3 ai 61 Cl bi bz 62 . 14. a2 62 C2 = - Cl C2 C3 ^3 bs C3 ai bi Cl as bz C3 a2 62 C2 0. 16. max 61 Cl ma2 62 C2 maz bz cz ax 61 Cl as &2 02 as 68 cs (ai + 61) 61 Cl («2 + 62) 62 C2 (as + bz) bz C3 ai bx Cl a2 62 C2 az bz cz 314. Inversions. Let us consider the permutations of a set of ob- jects, such as letters or numbers, and let us fix a certain particular order of the objects which we shall designate as the normal order. An inversion is said to occur in any permutation when an object is followed by one which in the normal order precedes it. Thus if abed is the normal order, then there are two inversions in bade. If 1234 is the normal order, then there are three inversions in 1432. Theorem. If in a given permutation, two objects are interchanged, the number of inversions with respect to the normal order is increased or decreased by an odd number. Let us consider the permutations Xrs Y and Xsr Y, where X and ;i" denote the groups of objects which precede and follow the interchanged objects r and s. Any inversion in Xand Fand any inversi(m due to the fact that X, r, s precede Y are common to Xrs Y and Xsr Y. Therefore, the number of inversions in Xrs Y is equal to the number in Xsr Y increased or decreased by 1 (according as rs is or is not in the normal order). Now let us consider two objects such as r and s separated by i objects. If the objects r and s are interchanged, the number of inversions is still changed by an odd number. For, by f + 1 interchanges of adjacent pairs the object r can be brought into the position immediately following s, and by i further interchanges of adjacent pairs, s may be brought to occupy XX, § 316] DETERMINANTS 483 the position formerly held by r. Each of these (i -f 1) + i = 2 i 4- 1 interchanges of adjacent pairs has increased or decreased the number of inversions by 1. Hence the net result of these 2^ + 1 interchanges has increased or decreased the number of inversions by an odd number. 315. Determinants of the nth Order. The square array (10) ai 6i a2 62 an K of n^ elements, such as we have considered for the cases n = 2, w = 3, is called a determinant of the nth order and will be denoted by the Greek letter A. This determinant will he understood to stand for the algebraic sum of all the different products of n factors each that can be formed by taking one and only one element from each row and one and only one element from each column, and giving to each such product a positive or negative sign according as the number of inversions of the subscripts {normal order 1, 2, •••, n) is even or odd, when the letters have the normal order ab -'-q. It should be noted that from the remarks in § 314 it follows that if we arrange the elements in any product so that the subscripts are in normal order, we can determine the sign of each term, by making it positive or negative according as the number of inversions of the letters is even or odd. 316. Properties of Determinants. Theorem 1. The expan- sion of a determinant of order n contains n ! terms. Proof. There are as many terms in the expansion of a determinant of the nth order as there are pennutations of the subscripts 1, 2, 3,---, n. But this number is n ! (§ 269) . Theorem 2. If each element of any row or column is multiplied by any constant m, the value of the determinant is multiplied by m. Proof. Since by the definition of a determinant, each term of the ex- pansion must contain one and only one element from each row and each .column, the factor m will appear once and only once in each term of the expansion. If m is factored out of this expansion, the remaining factor is the expansion of the original determinant. 484 MATHEMATICAL ANALYSIS [XX, § 316 Illustration. inaib2C3-\-ma2b3Ci+mazbiC2—maibzC2 — mazbiCz — mazbiPi moi bi Cl mctz b2 C2 = mt maz bz cz = m ai bi Cl 02 bz C2 az bz cz Theorem 3. The value of a determinant is not changed if roics and columns are interchanged, so that the first row becomes the first column, the second row the second column, and so on. This follows at once from the definition of the determinant and the paragraph immediately following it (§ 815) . Theorem 4. If two rows or two columns of a determinant are inter- changed, the sign of the determinant is changed. Illustration. See Ex. 14, p. 477, and Ex. 14, p. 482. Proof : Since by Theorem 3 rows and columns may be interchanged without affecting the value of the determinant, we need only consider the interchange of two rows. First, if two adjacent rows are interchanged, the order of the letters in the principal diagonal and in each term of the development is left unchanged. However two adjacent subscripts in each term of the expansion are interchanged, and hence the sign of every term is changed. Why ? Next consider the effect of interchanging two rows separated by k inter- mediate rows. By k interchanges of adjacent rows, the lower row can be brought just below the upper one. Now the upper row can be brought into the original position of the lower row Toy k -\- 1 further interchanges of adjacent rows. Therefore interchanging the two rows is equivalent to 2k-\-l interchanges of adjacent rows. But 2 A; + 1 is an odd number and therefore this process changes the sign of the determinant. Theorem 5. If two rows or two columns of a determinant are identical, the value of the determinant is zero. Proof : Let A be the value of the determinant and let the two identi- cal rows or columns be interchanged. Then, by Theorem 4, the value of the resulting determinant is — A. But since the rows or columns which were interchanged were identical, the value of the determinant is left unchanged. That is to say, A = — A or 2 A = 0, or A = 0. Corollary. If all the elements in any row or column are the same multiples of the corresponding elements in any other row or column, then the value of the determinant is zero. XX, § 317] DETERMINANTS 485 317. Minors. If we suppress the row and the column in which any given element appears, the determinant formed by the remaining elements is called the minor of that element. Illustration. In the determinant the minor of az is ax 6i Cl az &2 C2 «3 &3 cz and the minor of Cs is The minor of a\ is denoted by A\, of hj by ^y, etc. hi Cl bs Cs «i bi Cl2 62 EXERCISES 1. Prove that 2 2 3 1 1 5 4 4 9 = 0. 2. Prove that 3. Prove that 4 5 6 2 1 5 1 5 3 = 4 2 1 5 1 5 6 5 3 • 4. Prove that 3 4 5 2 4 1 8 4 5 ' = - 4 3 5 4 2 1 4 8 5 • 6. Prove that 4 1 3 -1 2 2 1 3 5 4 6 5 3 5 -1 -3 2-3 6 3 2 9 -1 1 12 = 0. 6. Prove that 26 9 28 18 30 3 < -5 10 25 = 30 13 14 15 3 -1 6 2 1 -5 3 5 8 1 1 2 5 =0. 2 4 10 7. How many inversions are there in the arrangement 4213765 if the normal order is 1234567 ? 8. How many inversions are there in the arrangement 46321 if the normal order is 42316 ? 486 MATHEMATICAL ANALYSIS [XX, § 317 9. Find the value of the minor of 5, of 6, of 7, for the determinant 4 5 1 3 6 2. 2 7 8 10. Write down the minor of as, of C2, of 64, for the determinant [1 &i ci di I2 62 C2 di 64 C4 d^ 11. Show that 1 2 5-1 3 3 6 2 4 2 7 3 5 15 4 1 6 4 3 2 5 12 3 5 7 6 = - 1 4 3 2 2 12 3 15 4 3 14 3 2 5 5 7 6 318. Additional Theorems. — The following theorems will be found useful in evaluating determinants. Theorem 6. Laplace's Expansion. If the product of each element in any row or column by its corresponding minor be given a positive or negative sign according as the sum of the number of the row and the num- ber of the column containing the element is even or odd, then the algebraic sum of these products is the value of the determinant. Proof : First, it is evident that in the development of the determinant, Ai is the coefficient of ai. For Ai is a determinant of order n — 1 in the elements a2, •••, a„, and its expansion contains a term for each permuta- tion of 2, 3, ••-, n. Moreover, the signs of the terms are correct ; for, the number of inversions is not changed by prefixing ai. Second, let us consider the element e situated in the zth row and the jth column. We can bring this element to the leading position, i.e. first row and first column, by i — 1 transpositions of rows and j — 1 transpositions of columns, i.e. hji-\-j — 2 transpositions in all. Therefore the sign of the determinant will have been changed i + j — 2 times. That is, if i + j is an even number, the sign of the determinant is left unchanged ; while if t+jis an odd number, the sign of the determinant is changed. Now that the element under consideration is in the leading position, we know from the first step that its coefficient is its minor. Since the relative posi- tions of the elements not in the ith row or the jth column are not effected by these transpositions, the minor of the element in its original position is the same as the minor of the element when it is in the leading position. XX, § 318] DETERMINANTS 487 Hence the coefficient of the element e, which is situated in the ith. row and the jth column, is (— ly+J E, where E is the minor of the element e. Corollary, If in the development of a determinant by minors with respect to a certain column (row) the elements of this column {row) are replaced by the corresponding elements of some other column {or row), the resulting expression vanishes. = aiAi — azAi + dsAs — a^A^i. We wish to show that, for example, 61^1 — 62^2 + 63^3 — b^Ai is zero. This expression is zero, for we have replaced the column of a's by the column of ?)'s and hence the determinant has two columns identical. The same proof applies to a determinant of order n. Theorem 7. If each of the elements of any row or column of a deter- minant consists of the sum of two numbers, the determinant may be expressed as the sum of two determinants. Proof : Let Illustration. ai bi Cl di a2 62 C2 d2 as 63 ca da a4 64 C4 d. (ai + a'l) (a2 + a'2) {an + a'„) 6, Qn be the given determinant. Expanding in terms of the first column we have (ai + a'i)Ai - (aa + ^'2)^2 + (aa + a'3)^3 +••• + (- 1)'*-K«» + «'n)^n or [aiAi — a^Ai + azA^ + ••• + (— l)"-ia„A] + [a'1^1 - a'2^2 + a'3^3 + ••• + (- l)"-ia'„J„], «1 bi ■ qi 02 62- ■'; + an bn- •Qn a'l Theorem 8. If to the elements in any row {or column) be added the corresponding elements of any other row {or column) each multiplied by a given number m, the value of the determinant is unchanged. The proof of this theorem follows easily from Theorems 7, 6, and 2. 488 MATHEMATICAL ANALYSIS [XX, § 319 319. The Evaluation of Determinants. We are now in a posi- tion to expand a determinant of any order. The following examples will illustrate the methods employed. Example 1. Expand A = Multiply the first column by columns. It gives A = 27 28 29 1 and add it to the second and third 25 26 26 27 27 28 25 1 2 26 1 2 27 1 2 By the corollary of Theorem 5, the value of this determinant is 0. Example 2. Expand the determinant 2-16 A = 1 14 6 3 4 2 7 4 3 12 5 We seek to transform this determinant in such a way as to make all the elements but one in some row or column 0. The second column looks most promising. We accordingly add 4 times the first row to the second row (this replaces the 4 in the second row by 0); we then add 2 times the first row to the third row (Why?) ; and then add the first row to the fourth row (Why ?) These operations give A = 2-1 5 1 9 26 7 8 17 6 5 7 6 9 26 7 — 8 17 6 5 7 6 — 2 26 7 2 17 6 1 7 6 The last determinant may be still further simplified as follows ; A = 2 26 7 2 17 6 -17 6 __ 140 191 -"131 18|- = _ (162-31): 40 19 31 18 -17 6 9 1 31 18 131. * This determinant is obtained from the preceding by subtracting the elements of the last column from those of the first. XX, § 319] DETERMINANTS 489 EXERCISES Evaluate the following determinants. 5. 14 13 — 121 17 16 17 . 25 24 -18 34 23 12 23 34 21 . 14 35 26 18 26 24 29 39 49 , 37 35 11 2 - -2 1 1 1 - -1 4 2 2 - -2 1 -1 • 2 1 -1 3 4 -2 6 4 - -3 8 -4 2 8 3 1 4 1 23 24 25 26 12 13 14 15 32 33 34 35 2 2 2 2 7. 8. b c + d c b + d d 6 + c 2 a a^ a + b ab 2b 62 abed 6x00 c y d z 10. 11. 1 1 1 a 6 c . a2 62 C2 1 1 1 1 a 6 c d a2 62 C2 (22 a8 68 C« (28 12. abed a 6 c (2 a — 6 c 5 a — 6 — c (2 13. Prove that if a determinant v^hose elements are rational integral functions of some variable, as y, vanishes when y = 6, then y — b is a factor of the determinant. [Hint : Use the corollary of theorem 6.] 14. Solve by factoring Examples 10, 11, 12. 15. Factor into two factors a b c b c a cab 16. Factor a a2 be b 62 ca c C2 ab 490 MATHEMATICAL ANALYSIS [XX, § 320 320. Solution of a System of Linear Equations. Suppose we have n linear equations in n unknowns and we desire their solution. Let the equations be (11) aixi H- 6ia;2 + Cia-3 + a^Xi -f biXi + C2Xs + + P\Xn = gi QnXl + \X2 + CnXs + ••• + PnXn= ^n Let A be the determinant of the cofficients of the unknowns, i.e. (12) A= ax 6i • .. px at 62 . •• Vi an K • •• Vn The determinant A is called the determinant of the system. Multiply the equations by ^1, — J.2, Az, — A4, etc., respectively, and add the re- sults. Then we have (13) Xi(aiAi — a^Az •••) + X2(hiA\ — h^Ai •••)+ •• + ic„(pi^i-p2^2---) = qiAi—qzAi •••• From the corollary of Theorem 6 it follows that the coeflBcient of x\ is A and that the coefficients of the other unknowns are zero. Moreover, the right-hand member of (13) is the expansion of A if we replace the column of a's by the column of constant terms. This determinant will be denoted by A^^. Therefore we may write or provided A rjt 0. Similarly provided A :?t 0. A . iCi = Ao A Xi -^ It will be noticed that this is a direct extension of the methods employed in §§ 311, 313. The result may be stated in words as follows. The value of any unknown is equal to a fraction whose denominator is the determinant of the system and whose numerator is the determinant obtained from the former by replacing the coefficients of the unknown sought by the column of constant terms. XX, § 322] DETERMINANTS 491 321. The Case A = 0. The previous methods show that, even if A = 0, we can derive from the given equations the relations A . a:i = Aog, A • X2 = A^g, •••, A • a:« = Apq. Now if A = 0, th6se relations would imply that Aa3 = 0, A5, = 0, ..., Ap5 = 0. But it is easy to write down a system in which A = and one or more of the Aog, Aftg««-are not zero. Such a system is then clearly inconsistent and has no solution. For example, 2 xi + X2 = 1, 2 Xi + a;2 = 2. If Aag = Afeg = ••• = Apg = 0, the system may be consistent but the un- knowns Xi, iC2, •••, Xn are not then completely determined. For example, 2 xi + 3^2 = 1, 4 jci + 2 X2 = 2. A complete discussion of this case is beyond the scope of this book.* 322. Consistent Equations. Equations which have a common solution are called consistent. Consider the three equations in two un- knowns X and y : (14) aix + biy -|- ci = 0. (15) a2X + h-iy + C2 = 0. (16) azx + hzy + Cg = 0. Two cases arise according as to whether a pair of the three equations has a single or an infinite number of solutions. Case 1. A single solution. In order that these three equations be con- sistent it is necessary that Cl h C2 &2 dl &1 a2 &2 (17) satisfy equation (16), i.e. that or its equivalent ai C\ «2 C'2 ai bi a2 &2 L|a2 62I J Cl bi Ol Cl ai 61 - as -63 + C3 C2 62 (H C2 aa &2 = = 0. * Those interested in this problem will find a complete discussion in BocHEB, Higher Algebra, Chapter IV. 492 MATHEMATICAL ANALYSIS [XX, § 322 Case 2. An infinite number of solutions. In this caae ttl _ (Z2 __ Cli bi 62 bs and hence, by the corollary of theorem 5, the above determinant must equal zero. Therefore, in order that three linear equations in two unknowns have a common solution, it is necessary that the determinant of the coeffi- cients of the unknowns and the known terms vanish. Extending this result to n linear equations in n — 1 unknowns, we have a necessary condition that n linear equations in n — 1 unknowns he consistent is that the determinant formed from the coefficients of the un- knowns and the knoion terms must vanish. It must be clearly understood that the vanishing of the above determi- nant is only a necessary and not a suificient condition that the equations be consistent. For example, the system 2x+ y-l = 0, 2x+ y + 5 = 0, 4a; + 2y + 3 = 0, gives 2 1-1 A= 2 1 6 =0, 4 2 3 but the equations are inconsistent, for any pair are inconsistent. EXERCISES Solve the following systems of equations by means of determinants ; 1. 8. < 2x — y — z = 0j Zx + y+z = 6, 2x-Sy — v=—2, 2x-hSv — d. — a; + ?/ + 2 = 2m, x-y + z=2n, x-\-y — z-2p. 32/ -4x- 20 + 10 =-21, a; + 72/ + 2-w> = 13, l/-2a;-30 + 2io = 14, .3a; + 5^ - 50 +3wj = 11. 4. x + y + w = Q, x + y + = 7, y + + w = 8, X + + to = 9. 05 + 2^-0 + 3to=- 10, x + 3y — 20 — 4to = l, 2x-y-30 + 5to = 3, 3x-.y-0-2io = 18. XX, § 322] DETERMINANTS 493 Determine whether the following systems of equations are consistent : 6. |a; + y-2 = 0, 7. ISx-2y + 'l=0. 8. lx-y-\-7=0, [4x+y-2 = 0. [3x + y-2 = 0. 3a;-6y-2=0. 9. Find k so that the following equations are consistent : 2x + y-3 = 0, 3ic-2/ = 2, x-^y + k = 0. / MISCELLANEOUS EXERf^ISES 1. Prove that the equation of the oifcle that passes through the points (aJi,yi), (2:2,2/2), (353,2/3) is (x2 + y2) X y \ (ici^ + yi^) a:i y-i 1 (X2^ + y'J^) ^2 2/2 1 (a;82 + 2/3^) a;3 2/3 1 = 0. 2. Prove that ax^ + two linear functions if + 2 hxy + 2fx + 2gy + c is the product of a h g h b f 9 f c 3. Prove that a necessary condition that the three lines aix 4- biy + Ci 0, a2X + 622/ + C2 = 0, asx + 632/ + ca = 0, be concurrent is that ai 61 Ci 052 62 C2 = 0. as 63 C3 Is this condition also suflBcient ? 4. Prove that the locus of the equation ax-^by + = is a straight line. [Hint : Let {xi ,2/1), (icz , 2/2) be any two fixed points on the locus and {x, y) any other point on the locus. Then we have axi + 6yi + c = 0, ax2 + by2 -h c = 0, ax + by -\- c = 0. Since these equations are consistent, the determinant of the coefficient is zero.] PART V. FUNCTIONS OF TWO VARIABLES SOLID ANALYTIC GEOMETRY CHAPTER XXI LINEAR FUNCTIONS THE PLANE AND STRAIGHT LINE 323. Introduction. Thus far the only functions which we have represented geometrically are those of the form y = f{x), i.e. functions of a single independent variable x. Such func- tions, in general, were seen to represent a curve in the (a;, y) plane. We shall now study functions of the form z = f(x, y), i.e. functions of two independent variables x and y. In order to carry out this investigation it is necessary to set up a coordi- nate system in three dimensions. 324. Orthogonal Projections. The orthogonal projection of a point P upon a plane a (Fig. 255) is the foot P' of the per- pendicular drawn from P to a. The ortho- gonal projection of a segment PQ upon a is the segment PQ' joining the projections of P and Q upon a. The orthogonal projection of a point P upon a line I is the foot P of the perpen- dicular drawn from P to I. The orthogonal projection of a segment PQ upon the line I is the segment PQ' joining the projections of P and Q upon I. 494 XXI, § 325] ^ LINEAR FUNCTIONS 495 325. Rectangular Coordinates in Space. Consider three mutually perpendicular planes intersecting in the lines X'X, F' F, ZiZ. These lines are themselves mutually perpendicular. The three planes are known as the coordinate planes and their three lines of intersection as the coordinate axes. The planes are known as the xy-plane, yz-plane, xz-plane, and the axes as the X-axis, y-axis, z-axis. The point which is common to the three planes and also to the three axes, is called the origin. The positive directions of these axes are usually taken as indicated by the arrows in Fig. 256. >-x Fig. 266 Fig. 257 Let P be any point in space, and let us consider the seg- ment OP. The numbers representing the projections of OP on the three axes we call the coordinates of P and denote them by x, y, and z. In Fig. 257, x = OA, y = OB, Z=:00. Conversely, any three real numbers x, y, z may be con- sidered as the coordinates of a point P. Why? If i^ is the foot of the perpendicular dropped from P on the xy-igleme, and A is the foot of the perpendicular dropped from Mon the a>axis, the coordinates of P are x — OA, y = AM, z = MP. The eight portions of space separated by the coordinate planes are called octants. From the preceding definitions it 496 MATHEMATICAL ANALYSIS [XXI, § 325 follows that the signs of the coordinates of a point P in any octant are as follows : (a) X is positive or negative according as P lies to the right or left of the 2/2!-plane ; (6) y is positive or negative according as P lies in front or back of the a;2;-plane ; (c) z is positive or negative according as P lies above or below the ajy-plane. EXERCISES 1. What are the coordinates of the origin ? 2. What is the z coordinate of any point in the a;y-plane ? 3. What are the x and y coordinates of any point on the 2:-axis ? 4. What is the locus of points for which x = ? for which y = ? for which ^ = ? 5. What is the locus of points for which x = and ?/ = ? 6. What is the locus of points for which y =0 and 2 = 0? 7. What is the locus of points for which s = and x = ? 8. What is the locus of points for which a; = 2 and y = 2 ? 9. If P(x, ?/, z) is any point in space, find (a) its distance from the xy-plane ; (d) its distance from the x-axis ; (h) its distance from the ?/2;-plane ; (e) its distance from the y-axis ; (c) its distance from the x^-plane ; (/) its distance from the ^-axis. 10. Describe the positions of each of the following points : (2, — 8, 3) ; (-2, 3, -5); (3, 3, -3); (-4, -7, -9). 11. Plot the following points : (2, 1, 3) ; (4, ~ 1, - 2) ; (0, 0, - 3) ; (3,1,1); (-1,-1,-1); (1,0,1); (-1,2,-1); (1,-1,0); (4, - 1, - 1). 12. Find the distance from the origin to the point P (a;, y, z). 13. A point P moves so that its distance from the origin is always equal to 4. Find the equation of the locus of P. 14. Show that the points (a;, j/, z^ and (— x, y, z) are symmetric with respect to the j/s-plane. 15. A rectangular parallelepiped has three of its faces in the coordi- nate planes. Find the coordinates of its vertices, assuming that the di- mensions of the parallelepiped are a, 6, c. XXI, § 326] LINEAR FUNCTIONS 497 Fig. 258 326. Directed Segments. We shall define the angle be- tween two directed lines I and m which do not meet, to be the angle between two similarly directed lines I' and m' which do meet (Fig. 258). Theorem I. If AB is a directed seg- ment on a line I, which makes an angle 6 with the directed line V, then (1) Proiv AB==AB cos Q. Proof : Through A' (Fig. 259) draw ?i parallel to I and let Bi be the projection of B on l^. Then, by definition, the angle between I and V is the same as the angle between l^ and V. It follows from § 135 *^^* A^B'==A'B, cose, or A'B' = AB cos By since A'Bi = AB. Theorem II. The projection on a directed line s of a broken line made up of the segments A1A2, AzA^, ^3^4? •••, A^^iA^, is equal to the projection on s of the segment AiA^. The proof of this theorem is left as an exercise. See § 136. Corollary. If Pi (%, yi, z^) and P^ {X2, y%y Zz) are any two points, then , ^ . 'X2 - xi = Proja, P1P2, (2) U2 - yi = Projy P1P2, ^2 - zi = Proj« P1P2. EXERCISES 1. Find the projections upon the coordinate axes of the sides of the polygon ABCDEF whose vertices are A (0, 0, 0), ^ (1, - 6, 4), C (- 2, 4, - 1), D (3, - 1, 2), E{2, 1, 4), F (1, 1, 1). 2. The projections of the segment MP upon the coordinate axes are 4, 3, - 1 respectively. If Jlf is (2, - 1, 3), find the coordinates of P. 2k 498 MATHEMATICAL ANALYSIS [XXI, § 327 327. Direction Cosines of a Line. Let I be any directed line and V a line through the origin having the same direction. If V makes angles a, y8, and y with the x, y, and z axes respec- tively, then, by definition, I makes the same angles with these axes. These angles are known as the direction angles of the line Z, while their cosines are called the direction cosines of I. Keversing the direction of a line changes the signs of the direc- tion cosines of the line. For reversing the direction of a line changes a, /?, y into rr — a, ir — (3, tt — y, respectively ; and by §122 cos(7r-^) = -cosa Theorem. The sum of the squares of the direction cosines of a line is equal to unity. )»X Fig. 260 Proof. Let P(x, y, z) be any point on V (Fig. 260). Then, we have {x = OP cos a, (3) 2/ = OP cos ^, [2; = OF cos y. Therefore, x^-j-y'' + z^= OP^[cos2 a + cos^ jS + cos^ y]. Since ips -f 2/2 -f gz = Op\* it follows that (4) cos2 a +C0S2 p 4- cos2 7 = 1. * or' = x2 + y2 and OP^ = 22 -I- oS^^ = ^2 -I- xa + ya. ± -\/P + m2 + n2 m ± VZ2 + 7^2 + n^' n , XXI, § 328] LINEAR FUNCTIONS 499 Any three numbers /, m, n, (not all zero) are proportional to the direction cosines of some line ; for, P{1, m, n) is a point and the direction cosines of OP are I cos a = (5) cos (3 cos y = ±VZ* + m2+n2 The direction cosines of OP are evidently proportional to Z, m, n, and they may be found by dividing I, m, and n, respec- tively, by ± V/2 + m^ 4- 7i\ 328. The Distance between Pi(xi, i^i, zj and ^2(^2, i/2, ^2). Let the direction angles of the segment P1P2 be a, (3, y. Pro- jecting P1P2 upon the axes, we have, from the corollary of § 326, P1P2 cos a = .T2 — Xi, P1P2 cos /? == 2/2 — 2/ij A A cos y = 22 — 2!l. Squaring and adding we have, by the theorem of § 327, i\P2"=(x2-x,y^(y2-yiy-h(z2-z,y. Therefore, (6) P,P2 = V(X2 - X,y + (1/2 - Vlf -h (^2 - ^l)^. EXERCISES 1. Find the length and the direction cosines of the segment P1P2, when (a) Pi is (2, 3, 4) and P2 is (- 1, 0, 5) ; (&) Pi is (- 1, 2, - 7), and P2 is (4, 1, 4) ; (c) Pi is (4, 7, 1), and P2 is (1, - 2, - 7). 2. Prove that the triangle whose vertices are A{m^ w, p), 5(n, p, m), C ( jo, m, n) is equilateral. 3. Find the direction cosines of a line which are proportional to 4, 7, 1. 4. Find the length of a line-segment whose projections on the co- ordinate axes are 4, 7, 2. 500 MATHEMATICAL ANALYSIS [XXI, § 329 329. The Angle between Two Directed Lines. If aj, p^, y^ and otg) ft, 72 ^^^ the direction angles of two directed lines li and ^2 the angle 6 between them may be determined as follows. Draw the lines I'l and I'z through the origin, parallel to the given lines (Fig. 261). Then the angle between l\ and V2 is 0. ^-x If P(x, y, z) is any point on l\, then, by Theorem II of § 326, we have ^^^.^^^ ^^ ^ p^^.^^^ OMNF, ^'^' OP cos e = OM cos a2 + MN cos ft + iVP cos y^. But, OiJf = OP cos «!, il[fiV = OP cos ft, NP = OP cos yj. Therefore, (7) cos e = COS tti cos a2 H- cos Pi cos P2 + cos 71 cos 72- We shall assume that 6 is the smallest positive angle satis- fying equation (7). 330. Parallel and Perpendicular Lines. If two lines are parallel and extend in the same direction, they are parallel to and agree in .direction with the same line through the origin. Therefore, if ai, ^1, yi and ag) ft? 72 a-^e the direction angles of the two lines, a^ = a^, /3i = ft, 71 = 72 ; and we may write (8) cos ai = cos a2, cos pi = cos p2, cos 71 = cos 72- Conversely, if relations (8) are satisfied, the given lines are parallel and extend in the same direction. Why ? XXI, § 331] LINEAR FUNCTIONS 501 If the two lines are parallel but extend in opposite directions, we have ai = ir — 03, ft^ tt — ft? yi = 'r — 72? and therefore, (9) cosai=-cosa2, cos Pi=-cos pg, cos 71 =- cos 72- Conversely, if relations (9) are satisfied, the given lines are parallel and extend in opposite directions. Why ? If the two lines are perpendicular^ it follows from formula (7) that, (10) cos tti cos ttg + COS pi COS p2 + COS 7i cos 72 = 0. Conversely, if (10) is true, the lines will be perpendicular. If /, m, n and l\ m', n' are proportional to the direction cosines of two lines, the lines will be perpendicular if, and only if, (11) W + mm' + nn' = 0. They will be parallel if, and only if, the numbers I, m, n are proportional to V, m', n'. If any of the numbers I, m, n are zero, the corresponding numbers of the set Z', m', n' must, of course, also be zero. 331. Point of Division. Let Pi(xi, 2/i> ^i), Afe Vii ^2) be two given points and P{x, y, z) any point on the segment P1P2 P P such that ^-^ = X. If a, jS, y are the direction angles of this PPi segment, it follows from § 326 that PiP cos a = x — aji, PP2 cos a= X2 — x. Therefore, PiP cos a __ X — Xi __, PP2 cos a X2^ X or Xi +Xx2 (12) x = 1-hX 502 MATHEMATICAL ANALYSIS [XXI, § 331 Similarly, we have It should be noticed that A is positive if P lies within the segment PiPo, and negative if it lies without. By varying A, the coordinates of any point (=^ P2) on the line P1P2 may be obtained. For the mid-point of P1P2 we have A = 1 and, hence, the coordinates of the mid-point of P1P2 are (14) ^^ a;i-f a;2 ^ y=yi±Jh^ ^^ ^1 + 22 ^ i^ J 2 EXERCISES 1. Find the cosine of the angle between the two lines whose direction cosines are proportional to 2, 3, 1 and — 1, 4, 6. 2. Find the coordinates of the points of trisection of the segment Pi(4, -1,3), P2(-4, 7,3). 3. Prove that the medians of the triangle whose vertices are (1, 2, 3), (3, 2, 1), (2, 1, 3) meet in a point. 4. Show that the following points are the vertices of a right triangle : (1,0,6), (7,3,4), (4, 5, -2). 6. If two of the direction angles of a line are 45° and 60°, find the third direction angle. 6. Prove that the values a = 30°, /3 = 30° are impossible. 7. The direction cosines of a line are m, 2 m, 3 m. Find wi. 8. Show that {x—\Y+{y-\- 2)2 -f. (5; _ 3)2 = 9 is the equation of a sphere whose center is at (1, — 2, 3) and whose radius is 3. 9. Express by an equation the fact that the point (x, y, z) is equi- distant from (2, 1, 3) and (— 1, 4, 3). 10. Show that the points (3, 7, 2), (4, 3, 1), (1, 6, 3), (2, 2, 2) are the vertices of a parallelogram. 11. Prove by two methods that the points (3, 6, 4), (4, 13, 3), (2, — 1, 6) are collinear. 12. Show that the points (4, 3,-4), (- 2, 3, 2), (- 2, 9, - 4) are the vertices of an equilateral triangle. XXI, § 332] LINEAR FUNCTIONS 503 13. Find the coordinates of the point which divides the segment Pi Pa in the ratio X, given (a) Pi(2, 6, 8), P2(-l, 3, 6),X = 3; (6) Pi(- 2, - 5, 8), P2(8, 0, - 2), X = - 2 ; (c) Pi(3, - 7, - 9), P2(2, - 2, - 1), X = 1. 14. Prove that the medians of the triangle Pi(a;i, yi, ^i), P2(X2, y2, ^2)1 Ps(x3, ys» Z3) meet in the point ( xi + ^2 + a;3 yi + y2 + Vs gi + g2 + ss \ V 3 ' 3 ' 3 / 15. Prove that the lines joining the mid-points of the opposite edges of a tetrahedron pass through a common point and are bisected by that point. 16. Are the following points collinear : (2, 1, 3), (—2, —5, 3), (1, 5, 7) ? 17. Find the direction cosines of the line that is equally inclined to the three axes. 18. Prove that the lines joining successively the middle points of the sides of any quadrilateral form a parallelogram. 19. Find the projection of the segment Pi(l, 2, 3), P2(2, 1, 3) upon the line that passes through the points P3(— 3, 5, — 5) , P4(8, — 9, 12). 332. Locus of an Equation. We saw that in the plane the locus of the equation f(x, y) =0 represents, in general, a curve. In an analogous way the equation f(x, y, z) = 0, in general, re- presents a surface. For, if we solve for z, we have z = F(x, y) and from this equation, we see that we can find, corresponding to every point (x, y) in the xy-^lame, one or more values of z (real or imaginary). The locus of the real points (x, y, z) is, in general, a surface, but may be a curve or a point. If there are no real values for a;, ?/, z which satisfy the equation f{x, y, z) — 0, we say that the equation has no locus. The locus of points satisfying the two conditions /(a;, y, 2!)=0 and F{x^ y^z)—^ is, in general, a curve in space, which is the intersection of the two surfaces represented by these equations. 504 MATHEMATICAL ANALYSIS [XXI, § 333 333. The Plane. A plane is defined as a surface such that every point collinear with two points of the surface is itself a point of the surface. We shall prove the following propositions : (a) Every equation of the first degree in x, y, and z represents a plane. ih) Every plane is represented by an equation of the first de- gree in X, y, z. To prove (a), let Pj (xi, yi, z^), P^ixz, 2/2? ^2) be any two points on the surface whose equation is Ax -{- By + Cz -\- D = 0. Then we have (15) Ax^ + By,+ Cz^ + D^(), (16) Ax^ 4- By2 -h Cz^ + 2) = 0. Now let Pi{x^, 2/3J 2:3) be any point on the line PiP^- Then (if P3 ^fc Pa)) there exists a value of A.(=5fc — 1) such that "^-"TTT' ^^-TTT' '^"TTX" ^^^^^^ We wish to show that the coordinates of this point also satisfy the equation Ax -\- By -\- Cz + D == 0. By substitution in this equation we have (17) .^-{Ax,-\- By^-\-Cz,-\-D)-^,:^^{Ax^-^By^^-Cz2+D) = 0. Relation (17) is true, since it follows from (15) and (16) that each parenthesis vanishes separately. Therefore the surface defined by the equation Ax -\- By ■\- Cz -\- D = satisfies the definition of a plane. To prove the statement (6), let tt be any plane, and let OH be the perpendicular from which meets ir in Pj (Fig. 262). The positive direction oi OR will be taken from to the plane. The direction angles of OR will be called a, )3, y XXI, § 333J LINEAR FUNCTIONS 505 and the length OPi will be denoted by p.* Now if P(Xj y, z) is any point in the plane, we have, by § 326, >3: Fig. 262 (18) Projo50P= Tvo]onOM+ ViojonMN-^ ProJo^iVP. Hence the equation (19) X cos a 4- y cos p + z cos y = p is the equation of the plane. Why ? It is seen to be an equa- tion of the first degree in x, y, z. This form of the equation is called the normal form. It follows from the above that, if Ax -^^ By -\- Cz -\- D = i^ the equation of a plane, the direction cosines of a line perpen- dicular to the plane are proportional to A, B, C. It is left as an exercise to prove that to reduce Ax -\- By -\- C^ -f- D = to the normal form we must divide each term by ± V^^ + B^ + C% the sign of the radical being chosen opposite to that oi D if D =^ 0, the same as that of C ii D = 0, the same as that of B if C=D = or the same as that of A if B=C=D = 0* * If the plane passes throilgh the origin we shall suppose OR is directed upward, and hence cos 7 >0 since 7 <7r/2. If the plane passes through the z-axis, then OR lies in the ccy-plane and cos7 = 0; in this case we shall sup- pose OR so directed that /3<7r/2 and hence cos/3>0. Finally if the plane coincides with the yz-plane, the positive direction on OR shall be taken as that on OX ^2 ±V^o^ B, ■-vc^' ±V^2^ + 02^ 506 MATHEMATICAL ANALYSIS [XXI, § 334 334. The Angle Between Two Planes. The angle between two planes is defined to be the angle between two normals {i.e. perpendiculars) to the planes. Let AiX -\- B^y + Cis; + Di = and A^x + Bay -\- C2Z -{- D2 = ^ be the equations of the two planes. The direction cosines of their normals are then (§ 333), cos «! = ^ , cos 02 = cos Bi = ^ cos ^2 = cos yi = ^ ; cos y2 = If ^ is the angle between these normals, then, from § 329, (20) cos ^ = ± A,A 2±BA + C,C2 V^l^ + B{^ + Ci2 V^z^ + ^2^ + C22 If the planes are perpendicular, cos ^ = 0, and we have (21) A^A^ + B1B2 4- Ci(72 = 0. If the planes are parallel their normals are parallel. Hence, by § 330, their equations in normal form are X cos a-\-y cosyS-j-g cos y—p, x cos «'+ ?/ cosy8'+2;cos y'=p', where either cos a = cos a', cos ft = cos ft', cos y = cos y', or cos a = — cos a', cos y8 = — cos ft', cos y = — cos y'. Therefore, if the two equations be written in the form A^x + Biy + Ciz + Di = 0, AzX + ^22/ + C^z + A = 0, . the planes will be parallel if and only if (22) A2 = kA,, B2 = kB,, Cz^kCi. (k^O) The equation of any plane parallel to Ax -\- By + Cz -\- D = can therefore be written in the form Ax -^ By + Cz + D' = 0. XXI, § 334] LINEAR FUNCTIONS 507 EXERCISES 1. Sketch the planes whose equations are (a) a; =2, (6) y = 4, {c) z=-6,(id) 2x + y = l, (e)y-z = 0. 2. How many arbitrary constants are there in the equation of the plane Ax+By+Cz + D = 0? 3. What is the general equation of a plane that passes through the origin ? 4. What is the equation of the x^z-plane ? y«-plane ? a;2!-plane ? 5. What are the intercepts on the axes of the planes whose equations are (o) 2x-3y-\-z = 12', (6) x-y+z=8; (c) x + y = 0; (d) 6x-7=0? 6. Give three numbers proportional to the direction cosines of the normal to the plane x-\-2y — z = 9. What are the direction cosines ? 7. What is the normal equation of the plane x — y + z = 9? 8. What is the equation of the system of planes parallel to 2x — y + 2! = 1? 9. What is the equation of the plane that passes through the origin and is parallel to 2x — Sy + 7z = b? 10. Show that the planes 2x + 4ty — z = 2 and 4a;— y-f40 = 7 are perpendicular. 11. What is the equation of the plane parallel to2x + 2y-\-z=:9 and 6 units farther from the origin ? 2 units nearer ? 12. What is the distance between the parallel planes 2x-i-2y + z = 9 and 2 X + 2 ?/ + « = 15 ? 13. Find the equation of the plane passing through the points (a) (1,2,1), (-1,1,0), (0,0,1); (&) (2,1,3), (1,1,2), (-1,1,4); (c) (2,2,2), (1,1, -2), (1,-1,0); (d) (1,1,-1), (1,-1,2), (-2,-2,2). [ Hint : Use the equation Ax-\-By+Cz + I)=0 and divide by any coeffi- cient that is not zero.] 14. If D 9^ show that the equation Ax -\- By + Cz ■}- D = can be written in the form x/a-{-y/b + z/c = l where a, b, c are the inter- cepts made by the plane on the x, y, z axes respectively. 15. Show that the four points (0, 0, 3), (4, —3, —9), (2, 1, 2), (4, 3, 3) are coplanar, i.e. they lie in the same plane. 508 MATHEMATICAL ANALYSIS [XXI, § 334 16. Find the equation of the plane that passes through the point P and is parallel to the plane a, when (a) Pis (2, 1, 8) and ais2£c + 3y- 50 = 5 ; (6) P is (1, 0, 0) and a is 2« + 2/ + = 1 ; (c) Pis (-2, - 1,6) and a is 3a;- 5 2/- 2« = 3. 17. Find the equation of the plane passing through the point P and perpendicular to the planes a and /3 when (a) P is (1, 1, 1), a is 2 x — 2/ — = 4, and ^isx-^y + z = l; (b) Pis (-1, 2, 1), a is a; + 2/ -3« = 3, and /3 is .Sx- 5y + 22 = 1 ; (c) Pis (0, 3, 4), ais2a: + 4y + = 7, and /3is2x-0 + 32! = 2. 18. Find the equation of the plane passing through the points Pi, Pa and perpendicular to the plane a, when (a) Pi is (1, 1, 1), P2is (-1,2, 1), and a is 2x- 3y - ;s = 2; (b) Pi is (0, 0, 1), P2 is (2, 1, 3), and aisx + y — 6;s = 0; (c) Pi is (2, 1, - 3), P2 is (0, 4, 2), and «is4x — y — ^ = 2. 19. Prove that the distance from the plane Ax + By + Cfe + D = to the point (XI, 2/1, 01) is + Ax, + By,+ Cz, + D^ VA^ + P2 + C=2 10. Find the distance from the plane a to the point P when (a) P is (2, 1, 4) and ais2x — iy -}■ z = 2 ; (6) Pis (2,3, - 1) and a is 2x + 2/ + 260-2 = 0; (c) P is (0, 0, 3) and ais3x-22/-50 = l. 21. Prove that the equation of the plane which passes through the point (xi, 2/1, «i) and is parallel to the plane Ax + By -\- Cz -{■ D = is A(x - xi) + B{3j - 2/1) + C{z - zi) = 0. 22. Prove that the equation of a plane which passes through the point (xi, 2/1, zi) and is perpendicular to the plane Ax+By+Cz+D = is (P01— C2/i)x+ (Cxi— ^01)2/+ (^2/1 — Bxi)z = 0. 23. Find the cosines of the angles between the following pairs of planes (a) 2x-32/ + = l, 2x + z=z0; (6) x-y-0 = 2, 2^-40 = 8; (c)x + = 3, 4x + 2/+30 = 5. 24. Find the equation of the plane that passes through Pi, P2 and makes an angle with the plane a, where (a) Pi is (0, -1, 0), P2 is (0, 0, -1), a is 2/+ 0-7 = 0, and d is 120°; (6) Pi is (1, 0, 1), P2 is (0, 1, 2), a is X + 2 2/ + 20 = 2, and 6 is 60°. XXI, § 335] LINEAR FUNCTIONS 509 25. Find the equation of the locus of a point which moves so that its distance from the x?/-plane is twice its distance from the a^-axis. 26. Find the equation of the locus of a point whose distance from the plane x + 2y — 5 = 0is twice its distance from the z-a,xis. 27. A point moves so that its distance from the origin is equal to its distance from the ^rx-plane. Find the equation of its locus. 335. Simultaneous Linear Equations. In § 70 we saw that three simultaneous linear equations in three unknowns have in general a single solution. We shall now show that three such simultaneous equations have either, (a) a single solu- tion, or (6) an infinite number of solutions, or (c) no solution. We shall prove this statement geometrically. Each equa- tion represents a plane ; the three planes may assume the fol- lowing relative positions. Case I. No two of the planes are parallel or coincident. (a) The three planes may intersect in a single point ; then there is a single solution of the three simultaneous equations. (&) The three planes may intersect in a line ; then there is an infinite number of solutions. (c) The three planes may intersect so that the three lines of intersection are parallel ; then there is no solution. Case II. Two of the planes are parallel but not coincident. In this case the three planes can have no point in common and the equations have no solution. Case III. Two of the planes are coincident. (a) The third plane may be parallel to the coincident planes, in which case there is no solution. (6) The third plane may intersect the coincident planes, in which case there is an infinite number of solutions. (c) The third plane may coincide with the coincident planes, in which case there is an infinite number of solutions. 510 MATHEMATICAL ANALYSIS [XXI, § 336 336. Pencil of Planes. All the planes that pass through a given line are said to form a pencil of planes. If A,x + B,y + Ciz H- A = 0, (23) ^ [A2X + B^y + C^z -f A = 0, are the equations of any two planes passing through the given line, then the equation of any other plane of the pencil can be written in the form (24) A^x + B^y j^C,z+D,+\ {A^x -h B^ + C<,z + A) = 0, where A is a constant whose value determines the particular plane of the pencil. (See § 68.) 337. Bundle of Planes. All the planes that pass through a common point are said to form a bundle of planes, and this common point is called the center of the bundle. If ' A^x + Biy+C^z + D^ = 0, (25) A2X + A2/ + C2Z + A = 0, A^x + B,y^C^z+ A-0, are the equations of any three planes passing through the center and not belonging to the same pencil, then the equation of any other plane of the bundle is (26) {A^x + Biy + C^z + D^) + X^A^x + B^y + C^z + A) + X2{Ax 4- B,y + C> + A) = 0, where Ai, A2 are constants whose values determine the position of the particular plane of the bundle. Why ? EXERCISES 1. Find the equation of the plane that passes through the Intersection of the planes a and /3 and the point P, when (a) ais2x + 3y--2; = l, /3isa; + 2/-2« = 2, and P is (1, 0, 2) j (6) aisx + y + 2^ = 0, /3is4a;-2?/-2; = l, and Pis (2, 1,1); (c) a is 3 a: - 2 y - 2r = 2, /3 is X - 2/ + « = 3, and P is (1, 0, 1). XXI, § 338] LINEAR FUNCTIONS 511 2. Show that the planes whose equations are 3a;--5y + 2 = 0, 6x + y = 2 2 + 13, lly — 2z = 17, belong to the same pencil. 3. What is the equation of the plane of the pencil whose axis is 2x — y + 52! + 2 = 0, 4x — Sy-\-z = l, which is perpendicular to the plane x = 0? y = 0? z = 0? 4. Find the equation of the plane that passes through the intersection of the planes 2x + y — z -j- 1, Sx— y — z = 2 and is perpendicular to the plane x -^ y — z = 1. 5. Find the equation of the plane that passes through the point of in- tersection of the planes a, /3, y and the points Pi, P2, when (a) ais2x + y = 1, fi is x — z = 1, yis2x — y + 2z = 3, Pi is (1, 0, 1), and P2 is (2, 1, 1) ; (b) aisSx- y -z = S, ^isx — y-\-2z = l, VisSx — 2y + 2! = 3, Pi is (2, 1, 3), and P2 is (0, 8, 0). 338. Equations of a Straight Line, (a) The two simul- taneous equations (27) ^ ^ ^ A2X + B^y-h C2Z -}- A = 0, represent a line, the intersection of the two planes, provided the two planes are not parallel. (6) A given point and a given direction determine a line. Let the given point be Pi{xi, y-^, z^ and a, /8, y the given direc- tion angles. If P (x, y, z) is any other point on the line at a distance d from Pj, then by § 326, dcos a=x—Xijd cos /8=2/— 2/ij d cos y = z — Zi. Hence we may write (28) x-x^ ^ y-yi ^ z- z^ ^ cos a cos p cos y * which are the equations of the required straight line. These equations are known as the symmetric equations of a straight line. In these equations cos a, cos /8, cos y can evidently be replaced by ar.y three numbers proportional to them. 512 MATHEMATICAL ANALYSIS [XXI, § 338 (c) Two distinct points Pi(xi,yi, Zi), P^ixz, 2/2? %) determine a line. Any line through, the point Pj is of the form cos a cos ^ cos y Now the direction cosines of PiA are proportional to a^ — ajj, ?/2 — J/i> 2^2 — 2:1. (§ 328.) Therefore the equations of the line through the points Pi, P.^ are (29) x-Xi ^ y -Vi ^ z-Zi X2 -Xi Vi- Vi ^2 - 2i We should note that in every case two equations are necessary to represent a line. Example 1. Reduce to the symmetric form the equations of the straight line, 2a: + y--0 = 3, x — y+2z-=:1. Eliminating y between the two equations we have 3 a; + = 10. Similarly, eliminating z we have 6 aj -f y = 13. Solving these two equations for x and equating the values found, we have a;_ y — 13 _ g-10 1~ -6 ~ -3 * The line is seen to pass through the point (0, 13, 10) and to have direction cosines proportional to 1, -- 5, — 3. Example 2. Find the equations of the line that passes through the point (4, —1, 3) and is perpendicular to the plane 2aj — 3y + 40=7. The required line is parallel to any line perpendicular to the plane and hence its direction cosines are proportional to 2, —3, 4 (§333). Therefore, the equation of the required line is a;~4 _ y + l _ g-3 2 -3 ~ 4 * EXERCISES 1. Write the equations of the line that passes through the point P and whose direction cosines are proportional to a, 6, c, where (a) Pis (1, 2, 1) and a = 2, 6 = - 7, c = 2; (6) P is (3, 0,-1) and a -2, 6 = 8, c = 9; (c) P is (3, - 2, -6) and a = 2, 6 = - 9, c = 3. XXI, § 338] LINEAR FUNCTIONS 513 2. Find the equations of the lines passing through the following pairs of points : (a) (2, 1, 4), (12, 2, 8) ; (c) (4, 3, 8), (8, - 2, 1) ; (6) (3, - 6, - 3), (- 3, 5, 7) ; (d) (5, 2, 1), (4, 7, -9). 3. Write in symmetric form the equations of the lines (a) 2x-7j + Sz = S, Sx + by -^z=:9; (b) Sx-y-z = 8, 4:X + Qy-5z = S; (c) 5x-\-Sy + z=S, 2x — y + z = 'l. 4. Find the equations of the line that passes through the point P and is perpendicular to the plane a, when (a) Pis (2, 1, 7) and ais3ic-?/ + 40 = 9; (6) P is (4, 2, - 2), and a is 2 a; - 6 2/ + 3 = 3 ; (c) Pis (— 1, 6, 3), and a is 3a; + 4?/ — ;s = 5. 6. Find the equations of the line that passes through the point (2,— 1, 4) and is parallel to the line x-_3_ y-7 _ g-7 4 ~ 2 -3 ' 6. Find in symmetric form the equations of the line that passes through the point (2, — 1, 4) and is parallel to the line 2x + y — z = 6, X — y + Zz = 4:. 7. Find the equation of the plane that passes through the point P and is perpendicular to the line Z, when (a) Pis(2,5, 1), andns^ = ^li = ^; o o o (6) Pis (- 1, 4,7), and I is =^ = 1^ = ^-±f. 8. Find the equation of the plane that passes through P(l, 5, 2) and is perpendicular to the line 3x-y+2 = 8,x — y + 22! = 6. 9 If tf is the angle between the two lines ^ ~ ^^ = V^rJll - lull^ «2 62 C2 10. Prove that the lines - = -^ = -^ , ~ = ^ = - are perpendicular to each other. 6-2-4463 2h CHAPTER XXII QUADRATIC FUNCTIONS. QUADRIC SURFACES 339. The Sphere. If a point P{x, y, z) moves so as to be always at a constant distance r (r > 0) from a fixed point {h, k, I), the locus of P is called a sphere. The equation of this locus is (1) (x-hy+(y-ky-{.(z-iy = r^. If this equation is expanded, it has the form (2) x^-\-y^-\-z^-\-Ax-{-By + Cz-^D = 0, where A, B, C, D are constants depending upon the coordinates of the center and the length of the radius. Conversely, an equation of the form (2), in general, repre- sents a sphere, for it can be written in the form <" ('+f)'-('-f)'H'+f)'=t'-f-?-^- which is a sphere if 4 4 4 The center of the sphere is at the point (— A/2, — B/2, — 0/2), and the radius is V^V4 + S2/4 + 074 - D. If the right-hand member of (3) is zero, the locus is the single point (— Ji/2, —B/2, — (7/2). If the right-hand member of (3) is less than zero, the equation has no locus. See § 206. 614 XXII, § 340] QUADRIC SURFACES 515 EXERCISES 1. Find the equation of the sphere whose center is at P and whose radius is r, when (a) Pis (2, 1, 9),andr = 6; (6) Pis(l, -8,0), andr = 2; (c) P is (4, -9,-2), and r = 7. 2. Find the equations of the eight spheres tangent to the three co- ordinate planes and having a radius of 4, 3. Find the equation of the sphere which has the line joining P(2, 6, 8) and ^(4, 6, 6) as a diameter. 4. Discuss the locus of each of the following equations, (a) x2 + ?/2 + ^2 _ 2 X - 2 y - 2 2 = 6 . (ft) a;2 + y2 _}. 2.2 _|. 4 a; __|_ 4 y _ 6 2 + 25 = 0. (c) a;2 + 2/2 + ^2 _ 2 X - 6y + 8 = 5. {d) x2 + 2/2 -f ;22 _ 2 2 - 4 y + 5 = 0. 5. Find the locus of points the ratio of whose distances from (0, 1, 0) and (1, 2, 3) is 5. 6. Show that the equation of the tangent plane to the sphere x'i + y^^Z^ =^2 at the point (xi, j/i, zi) is xxi -f yyi + zzi = r2. [Hint : The tangent plane is perpendicular to the radius.] 7. Find the equation of the sphere passing through the following four points. (a) (1,2,3), (3,1,0), (2,1,0), (3,4, 1). (6) (2, 1, 0), (- 1, - 1, 0), (3, 0, 2), (0, 0,0). [Hint : Use the equation x^ -h y^ -^ z'^ + Ax -]- By -\- Cz + D = and determine the values of A, J5, O, Z>.] 340. Cylinders. The surface generated by a straight line which moves parallel to a given line and always intersects a given fixed curve, is called a cylindrical surface or a cylinder. The generating line in any of its positions is called an element of the cylinder. Any algebraic equation in two cartesian coordinates repre- sents in space a cylinder whose elements are parallel to the axis of the third variable. 516 MATHEMATICAL ANALYSIS [XXII, § 340 For example, the equation a;2 + y2 =4 represents in the ccy-plane a circle (Fig. 263). But, the equation is satis- fied by the coordinates of any point P which lies on a line parallel to the «r 2;-axi8 and passes through a point Q on the circle. Moreover, if QP moves parallel to the 2-axis and continues to cut the circle the coordinates of P still satisfy the equation x^ + y2 _ 4, The cylinder traced by the line QP is the locus of the equation x^ + 2^2 _ 4, It is clear that if a cylinder has its axis parallel to a coordinate axis, a section made by a plane perpendicular to that axis is a curve parallel and equal to the directing curve on the coordinate plane. Thus the section cut by the plane z = S from the hyperbolic cylinder whose equa- tion is x^-y^ = 4, is a hyperbola equal and parallel to the hyperbola in the xy-plane whose equation is x^ — y^ = 4. 341. The Projecting Cylinders of a Curve. A cylinder whose elements are parallel to one of the coordinate axes and always intersect a fixed curve in space, is called a projecting cylinder of the curve. The equations of the projecting cylin- ders may be found by eliminating in turn each of the variables X, y, z, from the equations of the curve. Why ? The curve may often be constructed conveniently by means of two dis- tinct projecting cylinders. XXII, §342] QUADRIC SURFACES 517 EXERCISES 1. Describe the locus of each of the following equations, (a) aj = 2. (A) yz - 6. (6) 2x2 + y2 = 8. (4) ^_^ = o. ^^^ ^^ ^' (m) y2==3c2. (fir) 2/2 - ig2 = 1. 2. Prove that x^ + 2xy +y^ = 1 — z^ is the equation of a cylinder, the direction cosines of any element being proportional to (1, 1, 0). 3. Find the equations of the projecting cylinders of each of the following curves. Construct the curves as the intersection of two of these cylinders. (a) x2 + y2 + 2,2 = 4^ a;2 + y2 _ 5,2 = 0. (6) a; = 1, x2 + y2 4. ;5,2 = 4. {c) x^ - y^ = ^ Zj x^ + y^ = z. {d) y'2 = x + z,z = x + yK (e) 02 = xy, x2 = yz. 342. Symmetry, Intercepts, Traces, Sections. If a given equation is unaffected by replacing x hj — x throughout, the locus is symmetric with respect to the yz-^lane. If a given equation is unaffected by replacing y by — y, the locus is symmetric with respect to the a;2;-plane. If a given equation is unaffected by replacing zhj — z, the locus is symmetric with respect to the xy-i^\sine. What would be a test for symmetry with respect to the a;-axis ? the t/-axis ? the «-axis ? the origin ? The segments measured from the origin to where a surface cuts the axes are called the intercepts of the surface on the axes. To find the intercepts place two of the variables equal to zero and solve the resulting equation for the third variable. Why? 518 MATHEMATICAL ANALYSIS [XXII, § 342 The sections of a surface made by the coordinate planes are called the traces of the surface (Fig. 264). To find the equa- tions of the traces put each variable in turn in the given equation equal :to zero. Why? The equations f(x, y, z)—0 and x = k, a constant, are together the equations of the curve of intersection of the surface and a plane parallel to the i/^^-plane. Similarly sections parallel to the xy- and 2/z-planes may be found. If A: = 0, the sections are the traces. *-x 343. The Ellipsoid. The surface represented by the equation w 1' 4.1^' 4.^=1 a-i b^ c2 is called a,ii- ellipsoid. It is symmetric with respect to the three coordinate planes, the three axes, and the origin. The intercepts on the x-, y-, s-axes are respectively ± a, ±b, ± c (Fig. 266).* The traces on the three coordinate planes are, respectively. x^_^y &2 1, . = 0; ^' + ?-!=l, 2/ 0; Fig. 265 1, X = 0. The sections of the ellipsoid by the plane a; = A; is an ellipse whose equations are ('-'f<'-S) = 1, xz=k. bVl * The figure exhibits only that part of the surface lying in one octant,— that in which x, y, z are all positive. XXII, § 343] QUADRIC SURFACES 519 The semi-axes of this ellipse are 6 VI — W-ja^, cVT-^F/o^. As I A; I increases from to a, the axes of this elliptical section decrease. When | A; | = a the ellipse reduces to a point, and when I A; I > a the sections are imaginary. The surface lies therefore entirely between the planes x — a^ x = — a. Sim- ilarly it may be shown that the surface is also bounded by the planes y = b, y — — b ; z = c, z = — r. LI ' n --^ Fig. 26() A general idea of the appearance of an ellipsoid is given by Fig. 266, which represents a plaster model of this sur- face. Special Cases. In general the semi-axes a, b, c are unequal, but it may happen that two or three of them are equal. If the three are equal, i.e. a = b = c, the surface is a sphere. If two are equal, for example, if & = c, the ellipsoid is called an ellipsoid of revolution, for it can be generated by revolving the ellipse x'^/a^ -\- y^/b^ = 1, j; = about the avaxis. 520 MATHEMATICAL ANALYSIS [XXII, § 344 344. Surfaces of Revolution. The surface generated by revolving a plane curve about a line in its plane is called a surface of revolution. The equation of the surface is readily found when the axis of revolution, i.e. the line about which the curve is revolved, is one of the coordinate axes. Let y=f{x) be the equation of the plane curve in the a;?/-plane and the ic-axis the axis of revolution. As the curve 2/= /(a?) revolves about the a;-axis, any point P on this curve describes a circle, whose center is on the x-axis and whose radius is equal to f{x) (Fig. 267). Therefore for any position of P (x, y, z) we have, which is the equation of the required surface of revolution. If the ellipse rK'^/a^ -f y^/h^ = 1, ^ = is revolved about the x-axis, the equation of the surface of revolution is ^^ + ^^ = ita^-^^3, or 1 + 1^,4-^=1. EXERCISES 1. Sketch and discuss each of the following ellipsoida. (a) 9 a:2 + 4 2/2 + 16 z-^ = 144. (6) 25 a;2 + y2 _i_ ;j;2 =: loo. (C) x2 + 8y2 + 2 2;2=:16. 2. Show that the ellipsoid in Ex. 1 (6) is an ellipsoid of revolution. 3. Find the equations of the ellipsoids formed by revolving the follow- ing ellipses about the axes mentioned. (a) 9 x2 + 4 y2 = 36, « = o, x-axis. (6) 9 x2 + 4 y2 = 36, « = 0, 2/-axis. (c) 9 x2 + «2 = 9, ?/ = 0, ^-axis. (d) 25 2/2 + 4 22 = 100, X = 0, ?/-axi8. XXII, § 345] QUADRIC SURFACES 521 4. When an ellipse is revolved about its major axis the ellipsoid gen- erated is called a prolate spheroid; when it is revolved about its minor axis, an oblate spheroid. Which of the ellipsoids in Ex. 3 are oblate and which are prolate ? 5. Describe the locus of each of the following equations. (a) ^ + ^ + 5_ = o ^ ^ a^ b^ c'' ^ ^ a^ b^ d^ 345. The Hyperboloid of One Sheet. The surf ace. repre- sented by the equation (6) Qi "^ 62 ^2 is called a hyperboloid of one sheet respect to each of the coordinate planes, each of the coordinate axes, and the origin. The inter- cepts on the X- and ?/-axes are ± a and ± b respectively, while the surface does not meet the z-axis (Fig. 268). The traces on the coordinate planes are, respec- tively, It is symmetric with Fig. 268 --- = 1, 2/=0. a2 c2 ' ^ Of these, the trace on the xy-^lsme is an ellipse, while the other two are hyperbolas. The section of the surface made by the plane z = k, is an ellipse whose equations are •t'?]''t'n = 1, z = A;. 522 MATHEMATICAL ANALYSIS [XXII, § 345 This ellipse is real for all real values of k. The semi-axes are the smallest when A; = and increase without limit as | A: | increases. The plane y = X, | A. | ^ 6, intersects the surface in the hyperbola = 1, 2/ = X. •■['-^] I'-s] If I A I < 6 the transverse axis is parallel to the ic-axis, while if I A I > 6 it is parallel to the 2;-axis. Fig. 209 A good idea of the appearance of this surface is given by Fig. 269, which represents a plaster model of a portion of the surface. If A, = 6, the section consists of the two straight lines a c a c XXII, § 345] QUADRIC SURFACES 523 If A. = — 6, the section is the two lines a c a c These four straight lines lie entirely upon the surface. Similar considerations apply to the sections made by planes parallel to the yz-^lsme. . The form of one eighth of the surface is given in Fig. 268. The broken lines in that figure indicate three sections by the three planes y = \, for I A, I < 6, =b, and > b. Some of the straight lines on the surface are shown on the model represented by Fig. 269. If a = 6 the hyperboloid becomes a surface of revolution obtained by revolving the hyperbola x^/a^ — z^/c^ = 1, y =zO about its conjugate axis. EXERCISES 1. Sketch and discuss each of the following surfaces. (a) x2+4 2/2_g2=16. (ft) 9x^-\-y^-z^=Se. (c) ix^+16y^-z-^=Qi. 2. Are any of the surfaces in Ex. 1 surfaces of revolution ? 3. Show that (x-2y (y-l)2 (z-Sr ^^ 9 4 1 is the equation of a hyperboloid of one sheet whose center is at the point (2, 1, 3). 4. Show that ^-t: + ?l = i and -^ + l^+£!=3l are equations of a'^ b'^ c^ a'^ 52 ^2 hyperboloids of one sheet. 5. Find the equation of the hyperboloid of revolution formed by revolving each of the following hyperbolas about the axis specified. (a) 9 a;2 - 4 1/2 = 36, z = 0, transverse axis. (6) 9 x2 - 4 2/2 = 36, ;j; _ 0, conjugate axis. (c) 4 1/2 _ 2;2 _. 16, a; = 0, transverse axis. (d) 4 1/2. _ z2 = 16, X = 0, conjugate axis. 524 MATHEMATICAL ANALYSIS [XXII, § 346 346. Hyperboloid of Two Sheets. The surface represented by the equation (6) fl2 e,2 22 C2 ==1 is called a hyperboloid of two sheets. It is symmetric with ^x Fig. 270 respect to each of the coordinate planes, the coordinate axes and the origin. The intercepts on the a?-axis are ± a, while Fig. 271 the surface does not meet the y- or z-axis (Fig. 270). The traces on the coordinate planes are, respectively. a;2 y' There is no trace on the 2/2?-plane. XXII, § 346] QUADRIC SURFACES 525 The plane a; = A; intersects the surface in the curve whose equations, iik^± a, are 11-'] {5-] 1, x = k. If I A: I > a this curve is an ellipse ; if \k\ = a it is a point. If 1 A: I < a the equations have no locus. All sections parallel to the xy- and a;2;-planes are hyperbolas. A good idea of the appearance of this surface is given by Fig. 271, which represents a model of a portion of the surface. If 6 = c the hyperboloid becomes a surface of revolution formed by revolving the hyperbola X- about its transverse axis. EXERCISES 1. Construct and discuss each of the following surfaces, (a) 4 a;2 ~ 9 ^2 _ 36 ^2 = 144. (6) x2 - y2 _ z2 - 1. (c) 9x2-4«/2-02 = 36. 2. Are any»of the surfaces in Ex. 1 surfaces of revolution ? 3. Show that _ ±- + ^ - 5- = 1 and _ ±- _ ^^ + ^ = 1 are equations a2 &2 c2 02 62 ^c2 ^ of hyperboloids of two sheets. 4. Find the equation of the hyperboloid of revolution formed by revolving each of the following hyperbolas about the axis specified. (a) ^ - ^ = 1, = 0, conjugate axis. 4 9 (6) 4:y^ - z^=4, x = 0, transverse axis. (c) 2 x2 — 4 02 = 1^ y = 0, conjugate axis. 526 MATHEMATICAL ANALYSIS [XXII, § 347 347. The Elliptic Paraboloid. The surface represented by the equation is called an elliptic paraboloid. It is symmetric with respect to the xz- and 2/2j-planes, and the 2-axis. The intercepts on all three axes are zero. The trace on the a;y-plane is a point, namely the origin ; the traces on the xz- and yz-iplsmes are, respec- tively, the parabolas x^ = a^z, y = 0; y^ = b^z, x = (Fig. 272). Sections made by the planes z = k (k > 0) are ellipses. Why ? Those made by the planes x = k and y = kj respectively, are parabolas. Why ? Fig. 272 FiQ. 273 Figure 273 represents a model of a portion of the surface. If a = 6, the surface is a figure of revolution formed by re- volving the parabola x^ = a% y = 0, about the 2;-axis. XXII, § 348] QUADRIG SURFACES 527 348. The Hyperbolic Paraboloid. The surface represented by the equation x2 y2 (8) fe2 is called a hyperbolic paraboloid. (See Fig. 274.) It is sym- metric with respect to the xz- and yz-iplsmes. All three inter- cepts are zero. The trace on the xy-iplsme is the pair of lines ^ . ?/ 0, ^=0; the traces on the xz- and t/^-planes are, respectively, the parabolas x^ = a% y = 0; y^ = - b% a; = 0. Fig. 274 Fig. 275 Sections parallel to the a;?/-plane are hyperbolas, while those parallel to the xz- and 2/2!-planes are parabolas. The form of the surface is shown in Fig. ?75. 528 MATHEMATICAL ANALYSIS [XXII, § 348 EXERCISES 1. Sketch and discuss each of the following surfaces, (a) a;2 + 4y2 = 36 z. (c) y'^ - z^ = x. (6) 2X2 + ;32 = 16y. (d) 2X2 - 02 = _ y. 2. Sketch the surface «2-2a; + t^-4y = «~5. 349. The Cone. The surface represented by the equation is called a cone. It is symmetric with respect to the three coordinate planes, the three axes, and the origin. All three intercepts are zero. The trace on the xy- plane is a point, namely the origin. The traces on the x%- and 2/2!-planes are respec- tively the pairs of lines ca; ± az = 0, 2/ = ; c?/ ± 6^ = 0, a; = (Fig. 276). Sections paral- lel to the a;?/-plane are ellipses, while those parallel to the xz- and 2/2!-planea are hyper- bolas. If any point P (aji, ?/i, z^) on the sur- face is connected with the origin, then the line OP lies entirely on the surface. For, (Xaji, Xa:2, X%) are the coordinates of any point on this line (see § 331), and they arc seen to satisfy the given equation (9), for all values of X. If a = 6 the cone is a cone of revolution. EXERCISES 1. Construct and discuss each of the following surfaces. (a) x2 + t/2-;22 = o. (6)9x2 + 4 2/2-3602-0. (c) x2-y2 4.4;s2 = o. 2. A point P moves so as to be equidistant from a plane and a line perpendicular to the plane. Find the equation of the locus of P. 3. A point P moves so that the sum of its distances from the three coordinate planes is equal to its distance from the origin. Find the equation of the locus of P. XXII, § 350] QUADRIC SURFACES 529 350. Summary. The surfaces discussed are here enumer- ated for reference. Ellipsoid : 1 + !i + -2=l- (Figs. 265, 266, § 343) w- 0^ c^ HyPERBOLOID OF ONE SHEET: -! + S - ^ = ^- (^^^«- 2^^' 269, § 345) 0} ©2 Qi Hyperboloid of two sheets : a;2 ?/2 1. (Figs. 270, 271, § 346) Elliptic Paraboloid : a2 62 Hyperbolic Paraboloid : /2 t -I- 1 = z. (Figs. 272, 273, § 347) ^-|^ = ^. (Figs. 274, 275, § 348) a Quadric Cone: S + S-| = 0- (Kg. 276, § 349) QuADRic Cylinders : a;2 ±|-2 = 1, y^ = ^px, (§340) It is beyond the scope of this book to prove that the general equation of the second degree in three variables x, y, z, can, in general, be reduced to one of the above types. Those inter- ested in this problem will find it fully discussed in any stand- ard textbook on solid analytic geometry.* *See, for example, Snyder and Sisam, Analytic Oeometry of Space, Chapter 7. 2m 530 MATHEMATICAL ANALYSIS [XXII, § 351 y 351. Other Systems of Coordinates. Numerous systems of coordinates for deter- *'^ mining the position of a point P in space have been devised. The most common of these sys- • ^^ tems are the rectangular, polar, spherical, and cylindrical. A brief account of the last three systems follows. 352. Polar Coordinates. Consider the line OP drawn from the origin to any point P (Fig. 277). Let a, y8, y be the direction angles of OP, called the radius vector, and let p be the length of the radius vector. The four quantities a, p, y, p are called the polar coordinates of P. Conversely, any four quantities a, fi, y, p, with the restric- tion that cos2 a 4- cos^ ^ -)- cos^ y = 1, determine a point whose polar coordinates are a, p, y, p. Prove that tlie equations of transformation from rectangular to polar coordinates are. (10) x = p cos a, y = p cos p, z = p cos 7, p2 = x^^ +y^ -h z^ 353. Spherical Coordinates. Any point P in space de- termines (Fig. 278) the radius vector OP(=p), the angle <f> between the radius vector and the 2!-axis, and the angle 6 between the a>-axis and the pro- jection of the radius vector on the a;2^-plane. The quantities p, 0, <f> are called the spherical coordinates of the point P. The angle <^ is known as the colatitude, and the angle 6 as the longitude. Conversely, any three quantities p, 6, <f> determine in space a point P whose spherical coordinates are p, 6, <i>. Prove that the equations of transformation from rectangular to spheri- cal coordinates are, (11) X = pBind coB(p, y = pain6sm(f>, z — p cos 6. Fia. 278 XXII, § 354] QUADRIC SURFACES 531 354. Cylindrical Coordinates. Any point P in space de- termines (Fig. 279) its distance z from the ^p ajy-plane and the polar coordinates r, 6 of the point P' which is the projection of P on the a;?/-plane. These three quantities r, 0, z are called the cylindrical coordinates of P. Conversely, any three quantities r, B, z deter- mine a point whose cylindrical coordinates they are. Prove that the equations of transformations from rectangular to cylin- drical coordinates are, (12) x = rQ.os,e, ?/ = rsin^, z = z. EXERCISES 1. Express each of the following loci in spherical coordinates. (a) a;2 -I- ?/2 + z^ = 9. (6) a;2 _|. ^a _ 4^2 _ 0. (c) 4x2 + Qy'2 - z^ - 36. 2. Express each of the following loci in polar coordinates. (a) ic2 + 2/2 4. 2;2 = 16. (6) X + 1/ = 0. (c) 2 x^ - y'^ - ^^ = 0. 3. Express each of the following loci in cylindrical coordinates, (a) x^ + y'^ = 9. (6) x^ + y^ + z^ = 9. (c) z'^ - x'^ -\- y^ = 6. 4. Express the distance between two points in polar coordinates. 5. Find the polar, spherical, and cylindrical coordinates of the points whose rectangular coordinates are (2, 1, 4), (3, 3, 3). 6. What is the locus of points for which (a) ^ = a constant, = a constant (spherical coordinates) ? (6) r = a constant, 6 z= sl constant (cylindrical coordinates) ? 7. Find the general equation of a plane in polar coordinates. 8. Find the general equation of a plane in spherical coordinates; in cylindrical coordinates. 9. Show that in polar co5rdinates a point may be regarded as the intersection of a sphere and three cones of revolution which have an element in common. 10. Show that in spherical coordinates a point may be regarded as the intersection of a sphere, a plane, and a cone of revolution which are mutually perpendicular. 11. The spherical coordinates of a point are 5, 7r/4, 7r/6; find its rectangular coordinates ; its polar coordinates; its cylindrical coordinates. TABLES TO FOUE DECIMAL PLACES 534 Powers and Roots Squares and Cubes Square Roots and Cube Roots No. Sqitaee OUBK Square Root Cube Root No. Squabe Cttbe Square Root Cube Root 1 1 1 1.000 1.000 61 2,601 132,651 7.141 3.708 2 4 8 1.414 1.260 62 2,704 140,608 7.211 3.733 3 9 27 1.732 1.442 63 2,809 148,877 7.280 3.756 4 16 64 2.000 1.587 64 2,916 157,464 7.348 3.780 5 25 125 2.236 1.710 55 3,025 l(i6,375 7.416 3.803 6 36 216 2.449 1.817 56 3,136 175,616 7.483 3.826 7 49 343 2.646 1.913 57 3,249 185,193 7.550 3.849 8 64 512 2.828 2.000 58 3,364 195,112 7.616 3 871 9 81 729 3.000 2.080 59 3,481 205,379 7.681 3.893 10 100 1,000 3.162 2.154 60 3,600 216,000 7.746 3.915 11 121 1,331 3.317 2.224 61 3,721 226,981 7.810 3.936 12 144 1,728 3.464 2.289 62 3,844 238,328 7.874 3.958 13 169 2,197 3.606 2.351 63 3,969 250,047 7.937 3.979 14 196 2,744 3.742 2.410 64 4,096 262,144 8.000 4.000 15 225 3,375 3.873 2.466 65 4,225 274,625 8.062 4.021 16 256 4,096 4.000 2.520 66 4,356 287,496 8.124 4.041 17 289 4,913 4.123 2.571 67 4,489 300,763 8.185 4.062 18 324 5,832 4.243 2.621 68 4,624 314,432 8.246 4.082 19 361 6,859 4.359 2.668 69 4,761 328,509 8.307 4.102 20 400 8,000 4.472 2.714 70 4,900 343,000 8.367 4.121 21 441 9,261 4.583 2.759 71 5,041 357,911 8.426 4.141 22 484 10,648 4.690 2.802 72 5,184 373,248 8.485 4.160 23 529 12,167 4.796 2.844 73 5,329 389,017 8.544 4.179 24 576 13,824 4.899 2.884 74 5,476 405,224 8.602 4.198 25 625 15,625 5.000 2.924 75 5,625 421,875 8.660 4.217 26 676 17,576 5.099 2.962 76 5,776 438,976 8.718 4.236 27 729 19,683 5.196 3.000 77 5,929 456,533 8.775 4.254 28 784 21,952 5.292 3.037 78 6,084 474,552 8.832 4.273 29 841 24,389 5.385 3.072 79 6,241 493,039 8.888 4.291 30 900 27,000 6.477 3.107 80 6,400 512,000 8.944 4.309 31 961 29,791 5.568 3.141 81 6,561 531,441 9.000 4.327 32 1,024 32,768 5.657 3.175 82 6,724 551,368 9.055 4.344 33 1,089 35,937 5.745 3.208 83 6,889 571,787 9.110 4.362 34 1,156 39,304 5.831 3.240 84 7,056 592,704 9.165 4.380 35 1,225 42,875 5.916 3.271 85 7,225 614,125 9.220 4.397 36 1,296 46,656 6.000 3.302 86 7,396 636,056 9.274 4.414 37 1,369 50,653 6.083 3.332 87 7,569 658,503 9.327 4.431 38 1,444 54,872 6.164 3.362 88 7,744 681,472 9.381 4.448 39 1,521 59,319 6.245 3.391 89 7,921 704,969 9.434 4.465 40 1,600 64,000 6.325 3.420 90 8,100 729,000 9.487 4.481 41 1,681 68,921 6.403 3.448 91 8,281 753,571 9.539 4.498 42 1,764 74,088 6.481 3.476 92 8,464 778,688 9.592 4.514 43 1,849 79,507 6.557 3.503 93 8,649 804,357 9.644 4.531 44 1,936 85,184 6.633 3.530 94 8,836 830,584 9.695 4.547 45 2,025 91,125 6.708 3.557 95 9,025 857,375 9.747 4.563 46 2,116 97,336 6.782 3.583 96 9,216 884,736 9.798 4.579 47 2,209 103,823 6.856 3.609 97 9,409 912,673 9.849 4.595 48 2,304 110,592 6.928 3.634 98 9,604 941,192 9.899 4.610 49 2,401 117,649 7.000 3.659 99 9,801 970,299 9.950 4.626 60 2,500 125,000 7.071 3.684 100 10,000 1,000,000 10.000 4.642 For a more complete table, see The Macmillan Tables, pp. 94-111. Important Constants Certain Convenient Values for n = 1 to n 535 10 71 1/n Vn V^ n ! l/n\ LOGlO V 1 1.000000 1.00000 1.00000 1 1.0000000 0.000000000 2 0.600000 1.41421 1.25992 2 0.5000000 0.301029996 3 0.333333 1.73205 1.44225 0.1666667 0.477121255 4 0.250000 2.00000 1.58740 24 0.0416667 0.602059991 5 0.200000 2.23607 1.70998 120 0.0083333 0.698970004 6 0.166667 2.44949 1.81712 720 0.0013889 0.778151250 7 0.142857 2.64575 1.91293 5040 0.0001984 0.845098040 8 0.125000 2.82843 2.00000 40320 0.0000248 0.903089987 9 0.111111 3.00000 2.08008 362880 0.0000028 0.954242509 10 0.100000 3.16228 2.15443 3628800 0.0000003 1.000000000 Logarithms of Important Constants n =>= NUiMBBH A^ALUE OF n LoGio n TT 3.14159265 0.49714987 l-i-TT 0.31830989 9.50285013 7r2 9.86960440 0.99429975 V^ 1.77245385 0.24857494 e = Napierian Base 2.71828183 0.43429448 M= logio e 0.43429448 9.63778431 l-4-i»/=logelO 2.30258509 0.36221569 180 -J- TT = degrees in 1 radian 57.2957795 1.75812262 TT -M80 = radians in 1° 0.01745329 8.24187738 TT -4- 10800 = radians in 1' 0.0002908882 6.4(i372613 TT -T- 648000 = radians in 1" 0.000004848136811095 4.68557487 sin 1" 0.000004848136811076 4.68557487 tan 1" 0.000004848136811152 4.68557487 centimeters in 1 ft. 30.480 1.4840158 feet in 1 cm. 0.032808 8.5159842 inches in 1 m. 39.37 (exact legal value) 1.5951654 pounds in 1 kg. 2.20462 0.3433340 kilograms in 1 lb. 0.453593 9.65666(50 g (average value) 32.16 ft./sec./sec. 1.5073 = 981 cm./sec./sec. 2.9916690 -weight of 1 cu. ft. of water 62.425 lb. (max. density) 1.7953586 weight of 1 cu. ft. of air 0.0807 lb. (at 32° F.) 8.907 cu. in. in 1 (U. S.) gallon 231 (exact legal value) 2.3636120 ft. lb. per sec. in 1 H. P. 550. (exact legal value) 2.7403627 kg. m. per sec. in 1 H. P. 76.0404 1.8810445 watts in 1 H. P. 745.957 2.8727135 536 ) Four Place Logarithms N I 2 3 4 5 6 7 8 9 12 3 4 5 6 7 8 9 10 0000 0043 0080 0128 0170 0212 0253 02f)4 0334 0374 4 812 17 21 25 29 33 37 11 12 13 14 15 IG 17 18 19 0414 0702 1139 1461 1701 2041 2304 2553 2788 0453 0328 1173 1492 1790 2068 2330 2577 2810 0492 0864 1206 1523 1818 2095 2355 2.'>01 2833 0531 0899 1239 1553 1847 2122 2380 2625 2856 0569 0934 1271 1584 1875 2148 2405 2648 2878 0607 0969 1303 1614 1903 2175 2430 2672 2900 0645 1004 1335 1644 1931 2201 245". 2695 2923 0682 1038 1367 1673 1959 2227 2480 2718 2015 0719 1072 1399 1703 1987 2253 2504 2742 2967 0755 1106 1430 1732 2014 2279 2529 2765 2989 4 8 11 3 7 10 3 6 10 3 6 9 3 6 8 3 6 8 2 6 7 2 5 7 2 4 7 15 If, 23 14 17 21 13 16 19 12 16 18 11 14 17 11 13 16 10 12 16 9 1214 9 11 13 26 30 34 24 28 31 23 26 29 21 24 27 20 22 25 18 21 24 17 20 22 16 19 21 16 18 20 20 21 22 23 21 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 3010 3032 3054 3075 3096 3118 3139 3160 3181 3201 2 4 6 8 1113 16 17 19 3222 3424 3617 3802 3979 4150 4314 4472 4624 4771 4914 5051 5185 5315 5441 5563 5682 5798 5911 3213 3444 3636 3820 3997 4166 4330 4487 4639 4786 4928 5065 5198 5328 5453 5575 5694 5809 5922 3263 34G4 3655 3838 4014 4183 4346 4502 4651 4800 4942 5079 5211 5340 5465 5587 5705 5821 5933 3284 3483 3674 3856 4031 4200 4362 4518 4(569 3304 3502 3692 3874 4048 4216 4378 4533 4683 3324 3522 3711 3892 4065 4232 4393 4548 4698 3345 3541 3729 3909 4082 4249 4409 4564 4713 3365 3560 3747 3927 4099 4265 4425 4579 4728 3385 3579 3766 3946 4116 4281 4440 4594 4742 3404 3598 3784 3962 4133 4298 4456 4609 4757 2 4 6 2 4 6 2 4 6 2 4 5 2 4 6 2 3 6 2 3 5 2 3 6 13 4 8 10 12 8 10 12 7 9 11 7 9 11 7 9 10 7 8 10 6 8 9 6 8 9 6 7 9 14 16 18 14 16 17 13 16 17 12 14 16 12 14 16 11 13 16 11 12 14 11 12 14 10 12 13 4814 4955 5092 5224 5353 5478 5599 5717 6832 5944 4829 4843 4857 4871 4886 4900 1 3 4 6 7 9 10 11 13 4969 5105 5237 5366 5490 5611 5729 5843 6955 4983 5119 5250 5378 5502 6623 5740 5855 5966 4997 5132 6263 5391 5514 6635 6752 C866 6977 5011 6145 5276 6403 5527 6647 6763 5877 5988 6024 6159 6289 5416 5539 6658 5775 6888 5999 5038 5172 6302 6428 6551 6670 6786 5899 6010 1 3 4 1 3 4 13 4 1 2 4 12 4 12 4 12 4 1 2 3 1 2 3 6 7 8 5 7 8 5 7 8 6 6 8 6 6 7 5 6 7 5 6 7 6 6 7 4 6 7 10 11 12 911 12 9 1112 910 11 910 11 8 1011 8 9 11 8 9 10 8 910 40 6021 6031 6042 6053 6064 6075 6085 609(5 6107 6117 12 3 4 6 6 8 9 10 41 42 43 44 45 46 47 48 49 6128 6232 6335 6435 05 32 G628 (5721 6812 6902 6138 6213 6345 6444 6542 6637 6730 6821 6911 6149 6253 6355 6454 6551 664(5 6739 6830 6020 6160 6263 6365 6464 65(51 6656 6749 6839 6928 6170 6274 6375 6474 6571 6665 6758 6848 6937 6180 6284 6385 6484 6580 6675 6767 6857 6946 6191 6294 6395 6493 6590 (5684 6776 6866 6955 6201 6304 6405 6503 6599 6693 6785 6875 69(54 6212 6314 6416 6513 6609 6702 6794 6884 6972 6222 6325 6425 6622 6618 6712 6803 6893 6981 1 2 3 12 3 12 3 1 2 3 1 2 3 1 2 3 12 3 1 2 3 1 2 3 4 5 6 4 5 6 4 5 6 4 5 6 4 6 6 4 5 6 4 5 6 4 6 6 4 4 6 7 8 9 7 8 9 7 8 9 7 8 9 7 8 9 7 7 8 7 7 8 7 7 8 6 7 8 50 (5990 6998 7007 7016 7024 7033 7042 7050 7059 7067 12 3 3 4 6 6 7 8 51 52 53 54 7076 71G0 7243 7324 7084 7168 7251 7332 7093 7177 7259 7340 7101 7185 7267 7348 7110 7193 7275 7356 7118 7202 7284 7364 7126 7210 7292 7372 7135 7218 7300 7380 7143 7226 7308 7388 7152 7235 7316 739(5 1 2 3 1 2 3 12 2 1 2 2 3 4 5 3 4 5 3 4 5 3 4 6 6 7 8 6 7 7 6 6 7 6 6 7 N 1 2 3 4 5 6 7 8 9 12 2 4 5 6 7 8 9 The proportional parts are stated in full for every tenth at the right-hand side. The logarithm of any number of four significant figures can be read directly by add- ing the proportional part correspoTiding to the fourth figure to the tabular number correspondmg to the first three figures. There may be an error of 1 in the last place. 538 Four Place Trigonometric Functions [Characteristics of Logarith ms orni tted — determine by the usual rule from the value] Radiaks DT^aPTTfl Sine Tangent Cotangent Cosine i-.'£*ijri&x>i:<o Value Logio Value Logio Value Logio Value Logio .0000 .0029 0°00' 10 .0000 .0029 .0000 .0029 1.0000 .0000 90° 00' 1.5708 1.5679 .4637 .4637 343.77 .6363 1.0000 .0000 50 .0058 20 .0058 .7()48 .0058 .7648 171.89 .2352 1.0000 .0000 40 1.5650 .0087 30 .0087 .9408 .0087 .9409 114.59 .0591 1.0000 .0000 30 1.5621 .0116 40 .0116 .0658 .0116 .0658 85.940 .9342 .9999 .0000 20 1.5592 .0145 50 .0145 .1627 .0145 .1627 68.750 .8373 .9999 .0000 10 1.5563 .0175 1°00' .0175 .2419 .0175 .2419 57.290 .7581 .9998 .9999 89° 00' 1.5533 .0204 10 .0204 .3088 .0204 .3089 49.104 .6911 .9998 .9999 50 1.5504 .0233 20 .0233 .3668 .0233 .3669 42.964 .6331 .9997 .9999 40 1.547.-. .0202 30 .0262 .4179 .02()2 .4181 38.188 .5819 .99f)7 .9999 30 1.5440 .0291 40 .0291 .4637 .0291 .4638 34.368 .5362 .9996 .9998 20 1.5417 .0320 50 .0320 .5050 .0320 .5053 31.242 .4947 .9995 .9998 10 1.5388 .0349 2° 00' .0349 .5428 .0349 .5431 28.636 .4569 .9994 .99f)7 88° 00' 1.5359 .0.378 10 .0378 .5776 .0378 .5779 26.432 .4221 .9993 .9997 50 1.5330 .0407 20 .0407 .6097 .0407 .6101 24.542 .3899 .9992 .9996 40 1.5301 .0436 30 .0436 .6397 .0437 .6401 22.904 .3599 .99{)0 .9996 30 1.5272 .0465 40 .0465 .6677 .0466 .6682 21.470 .3318 .9989 .9995 20 1.5243 .0495 50 .0494 .6940 .0495 .6945 20.20(5 .3055 .t)988 .9995 10 1.5213 .0524 3° 00' .0523 .7188 .0524 .7194 19.081 .2806 .9986 .9994 87° 00' 1.5184 .0553 10 .0552 .7423 .0553 .7429 18.075 .2571 .9985 .9993 50 1.5155 .0582 20 .0581 .7645 .0582 .7652 17.169 .2348 .9983 .9993 40 1.5126 .0611 30 .0610 .7857 .0612 .7865 16..350 .2135 .9981 .9992 30 1.5097 .0640 40 .0640 .8059 .0641 .8067 15.605 .1933 .9980 .9991 20 1.5068 .0669 50 .0669 .8251 .0070 .8261 14.924 .1739 .9978 .9990 10 1.5039 .0698 4° 00' .0698 .8436 .0699 .8446 14.301 .1554 .9976 .9989 86° 00' 1.5010 .0727 10 .0727 .8613 .0729 .8624 13.727 .1376 .9974 .9989 50 1.4981 .0756 20 .0756 .8783 .0758 .8795 13.197 .1205 .9971 .9988 40 1.4952 .0785 30 .0785 .8946 .0787 .8960 12.706 .1040 .9969 .9987 30 1.4923 .0814 40 .0814 .9104 .0816 .9118 12.251 .0882 .9967 .9986 20 1.4893 .0844 50 .0843 .9256 .0846 .9272 11.826 .0728 .9964 .9985 10 1.4864 .0873 6° 00' .0872 .9403 .0875 .9420 11.430 .0580 .9962 .9983 86° 00' 1.4835 .0902 10 .0901 .9545 .0904 .9563 11.059 .0437 .9959 .9982 60 1.4806 .0931 20 .0929 .9682 .0934 .9701 10.712 .0299 .9957 .9981 40 1.4777 .0960 30 .0958 .9816 .0963 .9836 10.385 .0164 .9954 .9980 30 1.4748 .0985) 40 .0987 .9945 .0992 .9966 10.078 .0034 .9951 .9979 20 1.4719 .1018 50 .1016 .0070 .1022 .0093 9.7882 .9907 .9948 .9977 10 1.4690 .1047 6° 00' .1045 .0192 .1051 .0216 9.5144 .9784 .9945 .9976 84° 00' 1.4661 .1076 10 .1074 .0311 .1080 .0336 9.2553 .9664 .9942 .9975 50 1.4632 .1105 20 .1103 .0426 .1110 .0453 9.0098 .9547 .9939 .9973 40 1.4603 .1134 30 .1132 .0539 .1139 .0567 8.7769 .9433 .993(5 .9972 30 1.4573 .1164 40 .1161 .0648 .1169 .0678 8.5555 .9322 .<)i)32 .9971 20 1.4544 .1193 50 .1190 .0755 .1198 .0786 8.3450 .9214 .9929 .9909 10 1.4615 .1222 7° 00' .1219 .0859 .1228 .0891 8.1443 .9109 .9925 .9968 83° 00' 1.4486 .1251 10 .1248 .0961 .1257 .0995 7.9530 .JK)05 M^2 .9966 60 1.4457 .1280 20 .1276 .1060 .1287 .109(i 7.7704 .8904 .9918 .9964 40 1.4428 .1309 30 .1305 .1157 .1317 .1194 7.5958 .880(i .9914 .9^)63 30 1.4399 .1338 40 .1334 .1252 .1346 .1291 7.4287 .8709 .9911 .9961 20 1.4370 .1367 50 .1363 .1345 .1376 .1385 7.2(i87 .8615 .9907 .9959 10 1.4341 .1396 8° 00' .1392 .14.36 .1405 .1478 7.1154 .8522 .9903 .9958 82° 00' 1.4312 .1425 10 .1421 .1525 .1435 .1.569 6.9682 .8431 .9899 .9<)56 50 1.4283 .1454 20 .1449 .1612 .1465 .1658 6.8269 .8342 .9894 .9954 40 1.4254 .148'1 30 .1478 .1697 1495 .1745 6.6912 .8255 .9890 .9952 30 1.4224 .1513 40 .1507 .1781 .1524 .18.31 ().5606 .8169 .9886 .9950 20 1.41^)5 .1542 50 .1536 .1863 .1.5.54 .1915 6.4348 .8085 .9881 .9948 10 1.4166 .1571 9° 00' .1564 .1943 .1584 .1997 6.3138 .8003 .9877 .9946 81° 00' 1.4137 V^alne Logjo Value Lopin Value Lopio Value Logio Degrees Rapians Cosine Cotangent Tangent Sink Four Place Trigonometric Functions 539 [Charactoristics of Logarithms omitted — letermine by the usual rule from the value] Radians Degeees Sine Tangent Cotangent Cosine 7alue Logio Value Logio Value Logio V^alue Logio .1571 9° 00' .1564 .1943 .1584 .1997 6.3138 .8003 .9877 .9946 81° 00' 1.4137 .1600 10 .1593 .2022 .1614 .2078 6.1970 .7922 .9872 .9944 50 1.4108 .1029 20 .1622 .2100 .1044 .2158 6.0844 .7842 .98(38 .9942 40 1.4079 .1058 30 .1650 .2176 .1073 .2236 5.9758 .7764 .9863 .9940 30 1.4050 .1087 40 .1079 .2251 .1703 .2313 5.8708 .7687 .9858 .9938 20 1.4021 .1716 50 .1708 .2324 .1733 .2389 5.7694 .7611 .9853 .9936 10 1.3992 .1745 10° 00' .1736 .2397 .1763 .2463 5.6713 .7537 .9848 .9934 80° 00' 1.3963 .1774 10 .1765 .2468 .1793 .2536 5.5764 .7464 .9843 .9931 50 1.3934 .1804 20 .1794 .2538 .1823 .2609 5.4845 .7391 .9838 .9929 40 1.3904 .1833 30 .1822 .2600 .1853 .2680 5.3955 .7320 .9833 .992; 30 1.3875 .1802 40 .1851 .2674 .1883 .2750 5.3093 .7250 .9827 .9924 20 1.3840 .1891 50 .1880 .2740 .1914 .2819 5.2257 .7181 .9822 .9922 10 1.3817 .1920 11°00' .1908 .2806 .1944 .2887 5.1446 .7113 .9816 .9919 79° 00' 1.3788 .1949 10 .1937 .2870 .1974 .2953 5.0058 .7047 .9811 .9917 50 1.3759 .1978 20 .1965 .2934 .2004 .3020 4.9894 .6980 .9805 .9914 40 1.3730 .2007 30 .1994 .25)97 .2035 .3085 4.9152 .6915 .9799 .9912 30 1.3701 .2036 40 .2022 .3058 .2065 .3149 4.8430 .6851 .9793 .9909 20 1.3672 .2065 50 .2051 .3119 .2095 .3212 4.7729 .6788 .9787 .9907 10 1.3&43 .2094 12° 00' .2079 .3179 .2126 .3275 4.7046 .6725 .9781 .9904 78° 00' 1.3014 .2123 10 .2108 .3238 .2156 .3330 4.6382 .6664 .9775 .9901 ■50 1.3584 .2153 20 .2136 .3296 .2186 .3397 4.5736 .6603 .9769 .9899 40 1.3555 .2182 30 .2164 .3353 .2217 .3458 4.5107 .6542 .9763 .9896 30 1.3526 .2211 40 .2193 .3410 .2247 .3517 4.4494 .6483 .9757 .9893 20 1.3497 .2240 60 .2221 .3466 .2278 .3576 4.3897 .6424 .9750 .9890 10 1.3468 .2269 13° 00' .2250 .3521 .2309 .3634 4.3315 .636() .9744 .9887 77° 00' 1.3439 .2298 10 .2278 .3575 .2339 .3691 4.2747 .6309 .9737 .9884 50 1.3410 .2327 20 .2306 .3629 .2370 .3748 4.2193 .0252 .9730 .9881 40 1.3381 .2356 30 .2334 .3682 .2401 .3804 4.1653 .6196 .9724 .9878 30 1.3352 .2385 40 .2363 .3734 .2432 .3859 4.1126 .6141 .9717 .9875 20 1.3323 .2414 60 .2391 .3786 .2462 .3914 4.0611 .6080 .9710 .9872 10 1.3294 .2443 14° 00' .2419 .3837 .2493 .3968 4.0108 .6032 .9703 .9869 76° 00' 1.3265 .2473 10 .2447 .3887 .2524 .4021 3.9017 .5979 .96% .9866 50 1.3235 .2502 20 .2476 .3937 .2555 .4074 3.9136 .5926 .9689 .9863 40 1.3206 .2531 30 .2504 .3986 .2586 .4127 3.8667 .5873 .9681 .9859 30 1.3177 .2560 40 .2532 .4035 .2617 .4178 3.8208 .5822 .9674 .9856 20 1.3148 .2589 50 .2560 .4083 .2648 .4230 3.7760 .5770 .9(367 .9853 10 1.3119 .2618 15°00' .2588 .4130 .2679 .4281 3.7321 .5719 .9659 .9849 75° 00' 1.3090 .2647 10 .2616 .4177 .2711 .4331 3.0891 .5669 .9652 .9846 50 1.3061 .2676 20 .2644 .4223 .2742 .4381 3.6470 .5619 .9644 .9843 40 1.3032 .2705 30 .2672 .4269 .2773 .4430 3.6059 .5570 .9(>36 .9839 30 1.3003 .2734 40 .2700 .4314 .2805 .4479 3.5656 .5521 .9(528 .9836 20 1.2974 .2763 60 .2728 .4359 .2836 .4527 3.5261 .5473 .9621 .9832 10 1.2945 .2793 16° 00' .2756 .4403 .2867 .4575 3.4874 .5425 .9613 .9828 74° 00' 1.2915 .2822 10 .2784 .4447 .2899 .4622 3.4495 .5378 .9605 .9825 50 1.2886 .2851 20 .2812 .4491 .2931 .4669 3.4124 .5331 .9596 .9821 40 1.2857 .2880 30 .2840 .4533 .2962 .4716 3.3759 .5284 .9588 .9817 30 1.2828 .2909 40 .2868 .4576 .2994 .4762 3.3402 .5238 .9580 .9814 20 1.2799 .2938 50 .2896 .4618 .3026 .4808 3.3052 .5192 .9572 .9810 10 1.2770 .2967 17° 00' .2924 .4659 .3057 .4853 3.2709 .5147 .9563 .9806 73° 00' 1.2741 .2996 10 .2952 .4700 .3089 .4898 3.2371 .5102 .9555 .9802 50 1.2712 .3025 20 .2979 .4741 .3121 .4943 3.2041 .5057 .9546 .9798 40 1.2683 .3054 30 .3007 .4781 .3153 .4987 3.1716 .5013 .9537 .9794 30 1.2654 .3083 40 .3035 .4821 .3185 .5031 3.1397 .4969 .9528 .9790 20 1.2625 .3113 50 .3062 .4861 .3217 .5075 3.1084 .4925 .9520 .9786 10 1.2696 .3142 18^00' .3090 .4900 .3249 .5118 3.0777 .4882 .9511 .9782 72° 00' 1.2566 Value LoSio Value Loffio Value T^'^ffio Value Logjo Degeees Radians Co'JINE Cotangent Tangent Sine 540 Four Place Trigonometric Functions [Characteristics of Logarithms omitted — ietermine by the usual rule from the value] Kadians Deobees Sine Tangent Cotangent Cosine Value Logio Value Logio Value Logio Value Lopio .3142 18° 00' .3090 .4900 .3249 .5118 3.0777 .4882 .9511 .9782 72° 00' 1.2566 .3171 10 .3118 .4939 .3281 .6161 3.0475 .4839 .9502 .9778 60 1.2537 .3200 20 .3145 .4977 .3314 .6203 3.0178 .4797 .9492 .9774 40 1.2508 .3229 30 .3173 .5015 .3346 .6245 2.9887 .4756 .9483 .9770 30 1.2479 .3258 40 .3201 .5052 .3378 .6287 2.9600 .4713 .9474 .9765 20 1.2450 .3287 50 .3228 .5090 .3411 .5329 2.9319 .4671 .9465 .9761 10 1.2421 .3316 19° 00' .3256 .5126 .3443 .5370 2.9042 .4630 .9455 .9767 71° 00' 1.2392 .3346 10 .3283 .5163 .3476 .5411 2.8770 .4689 .9446 .9752 60 1.2363 .3374 20 .3311 .5199 .3508 .5451 2.8502 .4649 .9436 .9748 40 1.2334 .3403 30 .3338 .5235 .3541 .5491 2.8239 .4509 .9426 .9743 30 1.2305 .3432 40 .3365 .5270 .3574 .6531 2.7980 .4469 .9417 .9739 20 1.2275 .3462 50 .3393 .5306 .3607 .5571 2.7725 .4429 .9407 .9734 10 1.2246 .3491 20° 00' .3420 .5341 .3640 .6611 2.7476 .4389 .9397 .9730 70° 00' 1.2217 .3520 10 .3448 .5375 .3673 .5650 2.7228 .4350 .9387 .9725 50 1.2188 .3549 20 .3475 .5409 .3706 .5689 2.6986 .4311 .9377 .9721 40 1.2159 .3578 30 .3502 .5443 .3739 .6727 2.6746 .4273 .9367 .9716 30 1.2130 .3607 40 .3529 .5477 .3772 .6766 2.6511 .4234 .9356 .9711 20 1.2101 .3636 50 .3557 .5510 .3805 .5804 2.6279 .4196 .9346 .9706 10 1.2072 .3665 21° 00' .3584 .5643 .3839 .5842 2.6051 .4158 .9336 ,9702 69° 00' 1.2043 .3694 • 10 .3611 .5576 .3872 .6879 2.5826 .4121 .9325 .9697 60 1.2014 .3723 20 .3638 .5609 .3906 .5917 2.5605 .4083 .9315 .9692 40 1.1985 .3752 30 .3m5 .5641 .3939 .5954 2.5386 .4046 .9304 .9687 30 1.1956 .3782 40 .3692 .5673 .3973 .5991 2.5172 .4009 .9293 .9682 20 1.1926 .3811 50 .3719 .5704 .4006 .6028 2.4960 .3972 .9283 .9677 10 1.1897 .3840 22° 00' .3746 .6736 .4040 .6064 2.4751 .3936 .9272 .9672 68° 00' 1.1868 .3869 10 .3773 .5767 .4074 .6100 2.4545 .3900 .9261 .9667 50 1.1839 .3898 20 .3800 .5798 .4108 .6136 2.4342 .3864 .9250 .9661 40 1.1810 .3927 30 .3827 .6828 .4142 .6172 2.4142 .3828 .9239 .9(556 30 1.1781 .3956 40 .3854 .5859 .4176 .6208 2.3945 .3792 .9228 .9651 20 1.1752 .3985 50 .3881 .5889 .4210 .6243 2.3750 .3767 .9216 .9646 10 1.1723 .4014 23° 00' .3907 .6919 .4245 .6279 2.3559 .3721 .9205 .9640 67° 00' 1.1694 .4043 10 .3934 .5948 .4279 .6314 2.3309 .3686 .9194 .9635 50 1.1(>65 .4072 20 .3961 .5978 .4314 .6348 2,3183 .3652 .9182 .9629 40 1.1636 .4102 30 .3987 .6007 .4348 .6383 2.2998 .3617 .9171 .9624 30 1.160(5 .4131 40 .4014 .6036 .4383 .6417 2.2817 .3583 .9159 .9618 20 1.1577 .4160 60 .4041 .6065 .4417 .6462 2.2637 .3548 .9147 .9613 10 1.1548 .4189 24° 00' .4067 .6093 .4452 .6486 2.24(50 .3514 .9135 .9607 66° 00' 1.1519 .4218 10 .4094 .6121 .4487 .6520 2.2286 .3480 .9124 .9602 50 1.1490 .4247 20 .4120 .6149 .4522 .(5553 2.2113 .3447 .9112 .9596 40 1.1461 .4276 30 .4147 .6177 .4557 .6587 2.1943 .3413 .9100 .9590 30 1.1432 .4305 40 .4173 .6206 .4592 .6620 2.1775 .3380 .9088 .9584 20 1.1403 .4334 50 .4200 .6232 .4628 .6654 2.1609 .3346 .9075 .9579 10 1.1374 .4363 25° 00' .4226 .6259 .4663 .6687 2.1445 .3313 .9063 .9573 65° 00' 1.1345 .4392 10 .4253 .6286 .4699 .6720 2.1283 .3280 .<X)51 .95(57 50 1.1316 .4422 20 .4279 .6313 .4734 .6752 2.1123 .3248 .9038 .9561 40 1.128(5 .4451 30 .4305 .6340 .4770 .6785 2.0965 .3215 .9026 .95,^5 30 1.1257 .4480 40 .4331 .6.366 .4806 .6817 2.0809 .3183 .9013 .9549 20 1.1228 .4509 50 .4358 .6392 .4841 .6850 2.0656 .3150 .9001 .9543 10 1.1199 .4538 26° 00' .4384 .6418 .4877 .6882 2.0503 .3118 .8988 .9537 64° 00' 1.1170 .4567 10 .4410 .6444 .4913 .6914 2.0353 .3086 .8975 .9.530 50 1.1141 .4596 20 .4436 .6470 .4950 .6946 2.0204 .3054 .89(52 .9624 40 1.1112 .4625 30 .4462 .0495 .4986 .(5977 2.0057 .3023 .8949 .9518 30 1.1083 .4654 40 .4488 .6621 .5022 .7009 1.9^)12 .2991 .89:^5 .9512 20 1.1054 .4083 60 .4514 .6646 .5059 .7040 1.9768 .2960 .8923 .9505 10 1.1025 .4712 27° 00' .4540 .6670 .6096 .7072 1.9626 .2928 .8910 .9499 63° 00' 1.0996 Value Logio Value Lopio Value Logio Value Log,o Degrees Radians Cosine Cotangent Tangent Sine Four Place Trigonometric Functions 541 [Oharaoteristics of Logarithms omitted — determine by the usual rule from the value] Radians Dbgebes SI.B Tangent Cotangent Cosine Value Loffio Value Logio Value Logio Value Logio .4712 27° 00' .4540 .6570 .5095 .7072 1.9626 .2923 .8910 .9499 63° 00' 1.0996 .4741 10 .4566 .6595 .5132 .7103 1.9486 .2897 .8897 .9492 50 1.0966 .4771 20 .4592 .6620 .5169 .7134 1.9347 .2866 .8884 .948(5 40 1.0937 .4800 30 .4617 .6644 .5206 .7165 1.9210 .2835 .8870 .9479 30 1.0908 .4829 40 .4643 .6663 .5243 .7196 1.9074 .2804 .8857 .9473 20 1.0879 .4858 60 .4669 .6692 .5280 .7226 1.8940 .2774 .8843 .9466 10 1.0850 .4887 28° 00' .4695 .6716 .5317 .7257 1.8807 .2743 .8829 .9459 62° 00' 1.0821 .4Q16 10 .4720 .6740 .5354 .7287 1.8676 .2713 .8816 .9453 50 1.0792 .4945 20 .4746 .6763 .5392 .7317 1.8546 .2683 .8802 .9446 40 1.0703 .4974 30 .4772 .6787 .5430 .7348 1.8418 .2652 .8788 .9439 30 1.0734 .5003 40 .4797 .6810 .5167 .7378 1.8291 .2(522 .8774 .9432 20 1.0705 .5032 50 .4823 .6833 .5505 .7408 1.8165 .2592 .8760 .9425 10 1.0676 .5061 29° 00' .4848 .6856 .5543 .7438 1.8040 .2562 .8746 .9418 61° 00' 1.0647 .5091 10 .4874 .6878 .5581 .7467 1.7917 .2533 .8732 .9411 50 1.0617 .5120 20 .4899 .6901 .5619 .7497 1.7796 .2503 .8718 .9404 40 1.0588 .5149 30 .4924 .6923 .5658 .7526 1.7675 .2474 .8704 .9397 30 1.0559 .5178 40 .4950 .6946 .5696 .7556 1.7556 .2444 .8689 .9390 20 1.0530 .5207 50 .4975 .6968 .5735 .7585 1.7437 .2415 .8675 .9383 10 1.0501 .5236 30° 00' .5000 .6990 .5774 .761i 1.7321 .2386 .8660 .9375 60° 00' 1.0472 .5265 10 .5025 .7012 .7(544 1.7205 .2356 .8646 .9368 50 1.0443 .5294 20 .50jO .7033 .5851 .7673 1.7090 .2327 .8631 .9361 40 1.0414 .5323 30 .5075 .7055 .5890 .7701 1.6977 .2299 .8616 .9353 30 1.0385 .5352 40 .5100 .7076 .5930 .7730 1.6864 .2270 .8601 .9346 20 1.0356 .5381 50 .5125 .7097 .5969 .7759 1.6753 .2241 .8587 .9338 10 1.0327 .5411 31°00' .5150 .7118 .6009 .7788 1.6643 .2212 .8572 .9331 59° 00' 1.0297 .5440 10 .5175 .7139 .6048 .7816 1.6534 .2184 .8557 .9323 .50 1.0268 .5469 20 .5200 .7160 .6088 .7845 1.6426 .2155 .8542 .9315 40 1.0239 .5498 30 .5225 .7181 .6128 .7873 1.6319 .2127 .8526 .9308 30 1.0210 .5527 40 .5250 .7201 .6168 .7902 1.6212 .2098 .8511 .9300 20 1.0181 .5556 50 .5275 .7222 .6208 .7930 1.6107 .2070 .8496 .9292 10 1.0152 .5585 32° 00' .5299 .7242 .6249 .7958 1.6003 .2042 .8480 .9284 58° 00' 1.0123 .5614 10 .5324 .7262 .6289 .7986 1.5900 .2014 .8465 .9276 50 1.0094 .5643 20 .5348 .7282 .6330 .8014 1.5798 .1986 .8450 .9268 40 1.0065 .5672 30 .5373 .7302 .6371 .8042 1.5697 .1958 .8434 .9260 30 1.0036 .5701 40 .5398 .7322 .6412 .8070 1.5597 .1930 .8418 .9252 20 1.0007 .5730 50 .5422 .7342 .6453 .8097 1.5497 .1903 .8403 .9244 10 .9977 .5760 33° 00' .5446 .7361 .6494 .8125 1.5399 .1875 .8387 .9236 57° 00' .9948 .5789 10 .5471 .7380 .6536 .8153 1.5301 .1847 .8371 .9228 50 .9919 .5818 20 .5495 .7400 .6577 .8180 1.5204 .1820 .8.355 .9219 40 .9890 .5847 30 .5519 .7419 .6619 .8208 1.5108 .1792 .8339 .9211 30 .9861 .587(5 40 .5544 .7438 .6(561 .8235 1.5013 .1765 .8323 .9203 20 .9832 .5905 50 .5568 .7457 .6703 .8263 1.4919 .1737 .8307 .9194 10 .9803 .5934 34° 00' .5592 .7476 .6745 .8290 1.4826 .1710 .8290 .9186 66° 00' .9774 .5903 10 .5616 .7494 .6787 .8317 1.4733 .1683 .8274 .9177 50 .9745 .5992 20 .5640 .7513 .6830 .8344 1.4641 .1656 .8258 .9169 40 .9716 .6021 30 .5664 .7531 .6873 .8371 1.4550 .1629 .8241 .9160 30 .9687 .6050 40 .5688 .7550 .6916 .8398 1.4460 .1602 .8225 .9151 20 .9657 .6080 50 .5712 .7568 .6959 .8425 1.4370 .1575 .8208 .9142 10 .9628 .6109 35° 00' .5736 .7586 .7002 .8452 1.4281 .1548 .8192 .9134 55° 00' .9599 .6138 10 .5760 .7604 .7046 .8479 1.4193 .1521 .8175 .9125 50 .9570 .6167 20 .5783 .7622 .7089 .8506 1.4106 .1494 .8158 .9116 40 .9541 .6196 30 .5807 .7640 .7133 .8533 1.4019 .1467 .8141 .9107 30 .9512 .6225 40 .5831 .7657 .7177 .8559 1.3934 .1441 .8124 .9098 20 .9483 .6254 50 .5854 .7675 .7221 .8586 1.3848 .1414 .8107 .9089 10 .9454 .6283 36° 00' .5878 .7692 .7265 .8613 1.-3764 .1387 .8090 .9080 54° 00' .9425 Value Logio Value Loffio Value Logio Value I^Ogio Degeees Radians Cosine Cotangent Tangent Sine 542 Four Place Trigonometric functions [Characteristics of Logarithms omitted — determine by the usual rule from the value] Radians Deobeee 8lNB Tangent Cotangent Cosine Value Logio Value Logio Value Logifl Value Logic .6283 36° 00' .5878 .7692 .7265 .8613 1.3764 .1387 .8090 .9080 64° 00' .9425 .6332 10 .5901 .7710 .7310 .8639 1.3680 .1361 .8073 .9070 60 .9396 .6341 20 .5925 .7727 .7355 .86(J6 1.3597 .1334 .8056 .9061 40 .9367 .6370 30 .5948 .7744 .7400 .8692 1.3514 .1308 .8039 .9052 30 .9338 .6400 40 .5972 .7761 .7445 .8718 1.3432 .1282 .8021 .9042 20 .9308 .6429 50 .5995 .7778 .7490 .8745 1.3351 .1255 .8004 .9033 10 .9279 .6458 37° 00' .6018 .7795 .7536 .8771 1.3270 .1229 .7986 .9023 63° 00' .9250 .6487 10 .6041 .7811 .7581 .8797 1.3190 .1203 .79f39 .tX)14 50 .9221 .6516 20 .()065 .7828 .7627 .8824 1.3111 .1176 .7951 .9004 40 .9192 .6545 30 .6088 .7844 .7()73 .8850 1.3032 .1150 .7934 .8995 30 .9163 .6574 40 .6111 .7861 .7720 .8876 1.2954 .1124 .7916 .8985 20 .9134 .6603 50 .6134 .7877 .77(J6 .8902 1.2876 .1098 .7898 .8975 10 .9105 .6632 38° 00' .6157 .7893 .7813 .8928 1.2799 .1072 .7880 .8965 52° 00' .9076 .6601 10 .6180 .7910 .7860 .8954 1.2723 .1046 .7862 .8955 50 .9047 .6u90 20 .6202 .7926 .7907 .8980 1.2647 .1020 .7844 .8945 40 .9018 .6720 30 .6225 .7941 .79r>4 .9006 1.2572 .0994 .7826 .8935 30 .8988 .6749 40 .6248 .7957 .8002 .9032 1.2497 .0968 .7808 .8925 20 .8959 .6778 50 .6271 .7973 .8050 .9058 1.2423 .0942 .lim .8915 10 .8930 .6807 39° 00' .6293 .7989 .8098 .9084 1.2349 .0916 .7771 .8905 61° 00' .8901 .6836 10 .6316 .8004 .8146 .9110 1.2276 .0890 .7753 .8895 50 .8872 .6865 20 .6338 .8020 .8195 .9135 1.2203 .0865 .7735 .8884 40 .8843 .6894 30 .6361 .8035 .8243 .9161 1.2131 .0839 .7716 .8874 30 .8814 .6923 40 .6383 .8050 .8292 .9187 1.2059 .0813 .7698 .8864 20 .8785 .6952 60 .6406 .8066 .8342 .9212 1.1988 .0788 .7679 .8853 10 .8756 .6981 40° 00' .6428 .8081 .8391 .9238 1.1918 .0762 .7660 .8843 60° 00' .8727 .7010 . 10 .6450 .8096 .8441 .9264 1.1847 .0736 .7642 .8832 50 .8698 .7039 20 .6472 .8111 .8491 .9289 1.1778 .0711 .7623 .8821 40 .8668 .7069 30 .6494 .8125 .8541 .9315 1.1708 .0685 .7604 .8810 30 .8639 .7098 40 .6517 .8140 .8591 .9341 1.1640 .0659 .7585 .8800 20 .8610 .7127 50 .6539 .8155 .8642 .9366 1.1571 .0634 .7566 .8789 10 .8581 .7156 41° 00' .6561 .8169 .8693 .9392 1.1504 .0608 .7547 .8778 49° 00' .8552 .7185 10 .6583 .8184 .8744 .9417 1.1436 .0583 .7528 .8767 50 .8523 .7214 20 .6604 .8198 .87m .9443 1.1369 .0557 .7509 .8756 40 .8494 .7243 30 .6626 .8213 .8847 .94(38 1.1303 .0532 .7490 .8745 30 .8465 .7272 40 .6648 .8227 .8899 .9494 1.1237 .050(3 .7470 .8733 20 .8436 .7301 50 .6670 .8241 .8952 .9519 1.1171 .0481 .7451 .8722 10 .8407 .7330 42° 00' .6691 .8255 .9004 .9544 1.1106 .0456 .7431 .8711 48° 00' .8378 .7359 10 .6713 .8269 .9057 .9570 1.1041 .0430 .7412 .8699 50 .8348 .7389 20 .6734 .8283 .9110 .9595 1.0977 .0405 .7392 .8688 40 .8319 .7418 30 .6756 .8297 .91(33 .9()21 1.0913 .0379 .7373 .8676 30 .8290 .7447 40 .6777 .8311 .9217 .9646 1.0850 .0354 .7353 .8665 20 .8261 .7476 50 .()799 .8324 .9271 .9671 1.0786 .0329 .7333 .8(353 10 .8232 .7505 43° 00' .6820 .8338 .9325 .9697 1.0724 .0303 .7314 .8641 47°00' .8203 .7534 10 .6841 .8351 .9380 .9722 1.0661 .0278 .7294 .8629 60 .8174 .7563 20 .6862 .8365 .9435 .9747 1.0599 .0253 .7274 .8618 40 .8145 .7592 30 .6884 .8378 .9490 .9772 1.0538 .0228 .7254 .8606 30 .8116 .7621 40 .6905 .8391 .9545 .9798 1.0477 .0202 .7234 .8594 20 .8087 .7650 60 .6926 .8405 .9601 .9823 1.0416 .0177 .7214 .8582 10 .8058 .7679 44° 00' .6947 .8418 .9657 .9848 1.0355 .0152 .7193 .8569 46° 00' .8029 .7709 10 .6967 .»i31 .9713 .9874 1.0295 .012(3 .7173 .8567 60 .7999 .7738 20 .6988 .8444 .9770 .9899 1.0235 .0101 .7153 .8545 40 .7970 .7767 30 .7009 .8457 .9827 .9924 1.0176 .0076 .7133 .8532 30 .7941 .7796 40 .7030 .84(59 .9884 .9949 1.0117 .0051 .7112 .8520 20 .7912 .7825 50 .7050 .8482 .9942 .9975 1.0058 .0025 .7092 .8507 10 .7883 .7864 46° 00' .7071 .8495 1.0000 .0000 1.0000 .0000 .7071 .849.-. 45° 00' .7854 ' Value Logjo Value Lopio Value Logio Value Logio DSGSEEB Radians OOSINB Cotangent Tangent Sine INDEX Abscissa, 39. Absolute error, 236. Absolute value, of a directed quan- tity, 5 ; of a number, 36, 438. Addition, 41 ; graphic, 50 ; laws of, 52 ; of angles, 145 ; formulas, in trigonometry, 201 ; with rounded numbers, 238 ; of vectors, 435. Algebra, fundamental theorem of, 456. Algebraic functions, 143. Algebraic scale, 3, 5. Analytic geometry, 84. Analytic representation of a function, 21. Angle, definitions of, 143 ; directed — s, 143, 306 ; measurement of, 144, 188, 190; addition and subtraction of, 145 ; functions of, 147, 168; between lines, 146 306, 500; between planes, 506 of elevation and depression, 150 use of, in artillery service, 190 vectorial, 163 ; — s of triangle, 210; trisection of, 388; of a complex number, 438. Approximation, Newton's method of, 471. Arbitrary functions, 18, 19. Arc of a circle, 189. Archimedes, 386. Area, of a triangle, 186, 298 fif. ; of any polygon, 301. Arithmetic mean, 214. Arithmetic progression, 216. Arithmetic scale, 3, 5. Artillery service, use of angles in, 190. Asymptotes, 278, 280, 350. Axes, of reference, 38, 495 ; of ellipse, 273, 340 ; of hyperbola, 280, 349. Axioms, 53. Axis, of parabola, 109, 354; polar, 163 ; major and minor, 340 ; trans- verse, 280, 349 ; conjugate, 349. Binomial theorem, 428. Briggian logarithms, 227. Bundle of planes, 510. Cardioid, 383. Center, of pencil of lines, 91 ; of ellipse, 273, 282, 340; of circle, 283, 320 ; of hyperbola, 280, 349 ; of projection, 370; of bundle of planes, 510. Change, rate of, 68. Change ratio, 69. Characteristic of a logarithm, 227 ff. Circle, 320 ff . ; center and radius of, 320 ; cartesian equation of, 270, 320 ff . ; parametric equations of, 392 ; polar equation of, 381 ; as special case of an ellipse, 342 ; intersection of two — s, 330 ; or- thogonal — s, 331, 334; pencil of — s, 332 ; radical axis of, 332 ; radi- cal center of, 334 ; tangent to, point form, 324 ; slope form, 325 ; tan- gents from an external point to, 327; polar of point with respect to, 328 ; inversion with respect to, 336. Cissoid, 384. Cof unction, 177. Colatitude, 530. CoUinearity, condition for, 301. Combinations, 420 ff . Commensurable segments, 33. Common logarithms, 227. Completing the square, 113, 283 ff. 543 544 INDEX Complex numbers, 432 ff. ; defini- tion of, 432 ; geometric inter- pretation of, 433, 436 S. ; absolute value, angle, argument of, 438 ; polar form of, 438 ; multiplication and division of, 440; powers and roots of, 444. Complex roots of an equation, 462. Components of a vector, 436. Composite number, 414. Computation, numerical, 231, 236 ff., 242 ff. Conchoid, 385. Cone, 528. Conic or Conic section, 337 ff. ; as sections of a cone, 370. (See circle, ellipse, parabola, hyperbola.) Conjugate axis, 349. Conjugate complex numbers, 432. Conjugate diameters, 376. Conjugate hyperbolas, 353. Consistent equations, 94, 491. Constant function, 18. Continuous functions, 18, 102, 449. Coordinates, on a line, 37 ; rectan- gular, in a plane, 38 ; rectangular, in space, 495 ; polar, in a plane, 163, 377 ff. ; polar, in space, 530 ; spherical, 530; cylindrical, 531. Cosecant, 168; graph of, 173. Cosine, definition of, 147 ; variation of, 159 ; graph of, 159 ; graph of, in polar coordinates, 166 ; line representation of, 169 ; law of — s, 180 ; direction — s of a line, 498. Cotangent, definition of, 168; graph of, 173; line representation of, 169. Coversed sine, 168. Cube, duplication of, 388; table of — s, and — roots, 634. Cubic function, 129 ff. Cycloid, 399. Cylinders, 515. Cylindrical co5rdinates, 531. Decreasing function, 24. De Moivre's theorem, 443. Dependent variable, 12. Depressed equation, 469. Derivative, of a function, 451, 468 flF. ; successive — s, 458. Derived function, 451. Descartes's rule of signs, 466. Detached coefficients, 402 ff. Determinants, 475 ff. ; definition of, 475, 478, 483 ; evaluation of, 488 ; properties of, 483 ff, ; minor of, 485 ; Laplace's expansion of, 486 ; solution of equations by means of, 476, 480, 490. Diameter, of a conic, 373 ; conjugate — s, 376. Diodes, 384. Directed angles, 143, 306. Directed lines, 30G, 500. Directed quantities, 4. Directed segments, 5, 48, 497. Direction angles and cosines, 498. Directrix, of conic, 337 ; of ellipse, 340; of hyperbola, 350; of parab- ola, 355. Discontinuous functions, 18. Discriminant of a quadratic, 124. Distance, between two points in a plane, 294 ; in space, 499 ; of a point from a line, 311 ; of a point from a plane, 508. Division, by zero, 46 ; with rounded numbers, 241 ; of complex numbers, 440; point of, 295, 501 ; — transformation, 404. Duplication of cube, 388. Eccentricity, 337. Element of a determinant, 475, 478. Ellipse, definition of, 272, 338 axes and center of, 273, 340 equations of, 272, 282, 338 ff. slope of, 273 ; construction of, 346 ; focal radii of, 345 ; latus rectum of, 343 ; parametric equa- tions of, 347, 393 ; properties of, 340, 342, 365, 373; vertices, 343. Ellipsoid, 518 ff. Elliptic paraboloid, 526. Empirical function, 18. Equality, 51 ; conditional and un- conditional, 61. INDEX 545 Equation, definition of, 61 ; linear — s, 64, 83, 93 ff ., 509 ; quadratic, 120 £f.; trigonometric — s, 174; exponential — s, 234 ; solution by determinants, 476, 480, 490; in p-form, 467 ; depressed, 469 ; parametric, 392 ff. Error, absolute and relative, 236. Explicit function, 81. Exponential equations, 234. Exponential function, 217 ff. Exponents, 53, 218. External secant, 168. Factor, 51 ; of a polynomial, 404, 407; — theorem, 411. Factoring, solution of quadratic equation by, 121. Fire, indirect, in artillery service, 190. Focal radii, of ellipse, 345 ; of hyper- bola, 353. Focus, of conic, 337 ; of ellipse, 340 ; of hyperbola, 350; of parabola, 355. Forces, parallelogram of, 184, 435. Fractions, 33, 58 ; partial, 416 fif . Function, idea of, 1 ; definition of, 28 ; arbitrary, constant, empirical, 18; continuous, 18, 102, 449; representation of, 10, 13, 21, 22; increasing and decreasing, 24 ; linear, 64 ff., 494 ff. ; quadratic, 98 ff., 265 ff., 514 ff. ; cubic, 129 ff. ; power, 140; trigonometric, 147 ff., 168 ff.; logarithmic, 212 ff. ; ex- ponential, 217 ff. ; polynomial, 449 ff. ; explicit and implicit, 81 ; inverse trigonometric, 192 ff. ; sum of linear — s, 79 ; tables of — s, 534 ff. ; two- valued, 20, 265 ff. Functional notation, 409. Fundamental theorem of algebra, 456. Geometric mean, 214. Geometric progression, 216. Geometric representation, see graphic representation. Graphic addition, 7, 50. 2n Graphic interpolation, 13. Graphic representation, 3, 10, 37, 64, 433, 436 ff. Graphic solution of problems, 78, 123, 470. Graphs, statistical, 26 ; of linear functions, 72 ; of quadratic func- tions, 99 ff., 265 ff. ; of cubic functions, 129 ff. ; of polynomials, 452 ; of trigonometric functions, 158, 159, 161, 166, 173; of the exponential function, 221 ; of the logarithmic function, 224 ; in polar coordinates, 164, 378 ff. ; of parametric equations, 395. (.See entries under various classes of functions for further details.) Hesse's normal form of the equation of a straight line, 316. Highest common factor, 407. Hooke's law, 71. Hyperbola, definition of, 280, 338; axes and center, 280, 349 ; vertices of, 349; asymptotes of, 278, 280, 355; construction of, 354; equations of, 279, 283, 348 ff.; parametric equations of, 393 ; latus rectum of, 350 ; geometric properties of, 349, 350, 368; tangents and normals to, 359; conjugate — s, 353. Hyperbolic curves, 140. Hyperbolic paraboloid, 527. Hyperbolic spiral, 386. Hyperboloid, of one sheet, 621 ; of two sheets, 524. Hypocycloid, 400. Identities, definition of, 61 ; trigono- metric, 171. Imaginary number, 432 ff. Implicit functions, 81 ; quadratic functions, 265 ff. Incommensurable quantities, 34. Increasing function, 24. Independent variable, 12. Infinity, 48. ■ Initial line, 163. Inscribed circle, 187. 546 INDEX Integer, 33 ; properties of, 414. Intercept, 74, 517 ; — form of equa- tion of straight line, 315. Interpolation, 13, 77, 230. Intersection, of two lines, 93 ; of two circles, 330; of a line and a conic, 357. Inverse trigonometric functions, 192 ff. Inversion, with respect to a circle, 336 ; of order, 482. Inversor of Peaucellier, 336. Irrational numbers, 34 ; as roots of an equation, 470. Laplace's expansion of a determi- nant, 486. Latus rectum, of an ellipse, 343 ; of a hyperbola, 350 ; of a parabola, 355. Law, of signs, 49, 466; — s of ex- ponents, 53, 220; of sines, 180/ of cosines, 180 ; of tangents, 209. Less than, 39. Limafon, 383. Line representation of trigonometric functions, 169. Linear equations, 64, 83, 93, 476, 480, 490, 509. Linear functions, 64 fif., 494 ff. Linear interpolation, 77. Locus, of equation in rectangular coordinates, 67, 83, 509 ; in polar coordinates, 377. Logarithm, definition of, 223 ; in- vention of, 212; laws of, 225; systems of (natural and common), 226, 227 ; characteristic and man- tissa of, 227 ff. ; use of tables of, 229; — s in computation, 231, 242 £F.; tables of, 536 ff. Logarithmic paper, 260 ff. Logarithmic scale, 252. Logarithmic spiral, 387. Longitude, 530. Magnitude, 4. Major axis of ellipse, 340. Mathematical analysis, 30. Maxima and Minima, 109, 137, 453, 455. Measure, unit of, 2. 33, 34, 144, 188. Menelaus, theorem of, 319. Midpoint of a segment, 295, 502. MU, 190. Minor of a determinant, 485. Minor axis of ellipse, 340. Multiple roots, 460. Multiple-valued function, 20, 265 ff. Multiplication, 44, 50, 52, 239, 440. Napier, J., 212. Natural logarithms, 226. Newton's method of approximation, 471 ff. Nicomedes, 385. Normal, to a curve, 359 ; — form of the equation of a straight line, 316 ; of a plane, 505. Number, 2, 33 ; real, 3, 36 ; rational and irrational, 34 ; positive and neg- ative, 3; complex (imaginary), 432 ff. Octant, 495. One-valued function, 20. Ordinate, 39. Origin, 38, 163, 495. Orthogonal circles, 331, 334. Orthogonal projection in space, 494. Parabola, 354 ff. ; definition of, 109, 268, 338; equations of, 109, 282, 354 ff.; properties of, 355, 362; slope of, 268 ; tangents and normals to, 359, 362 ; latus rectxim of, 365. Parabolic curves, 140. Parabolic reflector, 364. Parabolic spiral, 389. Paraboloid, of revolution, 364 ; elliptic, 526; hyperbolic, 527. Parallel lines, in plane, 85 ; in space, 500. Parameter, definition of, 90, 392; of system of lines, 90. Parametric equations, 392 ff. Partial fractions, 416 ff. Pascal, B., 383,431 ; — 's triangle, 431. Peaucellier, inversor of, 336. Pencil, of lines, 91; of circles, 332; of planes, 510. INDEX 547 Period of trigonometric functions, 157, 159, 162. Permutations, 420 ff. Perpendicular lines, in a plane, 86 ; in space, 500. Plane, 504 ff. ; vectors in a, 435. Point of division, in a plane, 295 ; in space, 501. Polar and Pole, with respect to a circle, 328 ; with respect to a conic, 364, 373. Polar axis, 163. Polar coordinates, in a plane, 163 ff., 377 ff. ; in space, 530. Polar form of complex number, 438. Polygon, area of, 301. Polynomials, 402, 449 ff. Power function, 140. Powers, 53, 218, 444 ; table of, 534. Prime number, 414. Principal diagonal of a determinant, 475, 478. Principal value of inverse trigono- metric function, 193 ff. Probability, 426 ff. Product formulas in trigonometry, 207. Progression, arithmetic and geomet- ric, 216. ProjectUe, 397 ff. Projecting cylinder of a curve, 516. Projection, 196 ff. ; central, 370; orthogonal in space, 494, 497. Pure imaginary number, 432. Quadrant, 39 ; angles in a, 145. Quadratic equation, 120 ff. Quadratic function, 98 ff. ; applica- tions of, 115 ff.; slope of, 274, 287 ; implicit, 265 ff., 514 ff. Quantity, 2, 4 ; directed, 4. Radian, 188. Radical axis and center of two circles, 332, 334. Radius vector, 163, 530. Range, of variable, 23; of a pro- jectile, 397. "■ Ratio, 33 ; simple, 294, 319. Rational numbers, 33 ; as roots of an equation, 467. Reciprocal spiral, 386. Rectangular coordinates, in a plane, 38 ff.; in space, 495. Reference, axes of, 38. Reflector, parabolic, 364. Relative error, 236. Remainder theorem, 411. Revolution, surfaces of, 364, 519, 520. Roots, of numbers, 444 ; table of, 534 ; of an equation, 120, 455 ff. ; equal, of a quadratic, 124 ; rela- tion of, to coefficients, 125, 473 ; complex, 462 ; rational, 467 ; irrational, 470 ; multiple, 460 ; Newton's method of approximation to, 470 ff. Rotation in a plane, 200. Rounded numbers, 236 ff. Scale, arithmetic and algebraic, 3, 5 ; rectilinear, uniform and non- uniform, 5 ; logarithmic, 252. Secant, definition of, 168 ; graph of, 173 ; line representation of, 169. Section of a surface, 517. Segment, directed in a plane, 5, 48 ; in space, 497 ; — s to represent statistical data, 6. Shear, 132, 288. Significant figures, 236. Signs, law of, 49. Simple ratio, 294, 319. Simultaneous equations, 94 ff., 476, 480, 490. Sine, definition of, 147 ; variation of, 157 ; graph of, 158. Single-valued function, 20. Slide rule, 252 ff. Slope, 73, 135, 268, 273, 285, 287, 290, 449. Solution, of quadratic equations, 120 ff. ; of algebraic equations, 468 ff. ; of trigonometric equations, 174; of exponential equations, " 234 f of triangles, 181 ff., 244 ff. Sphere, 514. 548 INDEX Spherical co5rdinates, 530. Spirals, 386 ff. Statistical data and graphs, 6, 26. Straight line, 64 ff., 500 ff. ; slope of, 73; equations of, 83, 307 ff., 511; intercept form of, 89, 315; normal form of, 315 ; parallel and perpendicular — s, 85, 500 ; system of — s, 90 ; intersection of — s, 93 ; polar equation of, 380 ; pencil of — s, 91 ; direction cosines and angles of, 498. Subnormal, 364. Successive derivatives, 458. Sum of linear functions, 79. Surd, 35. Surface, 504 ff., 514 ff. ; of revolu- tion, 364, 519 ff. Symmetric equations of straight line, 511. Synthetic division, 412. Table, of squares, etc., 118, 534 ; of logarithms, 536 ff. ; of trigono- metric functions, 538 ff. Tabular representation of a function, 13. Tangent (to a curve), definition of, 103 ; to a circle, 324 ff. ; to a conic, 360 ff.; slope of, 73, 107, 135, 268, 273, 285, 287, 290, 449 ; slope forms of equation, 359 ; point forms of equation, 361. Tangent (trigonometry) , definition of, 147; variation of, 160; graph of, 161 ; line representation of, 172 ; law of — s, 209. Taylor's theorem, 458. Term, 51. Trace of a surface, 517. Transverse axis of a hyperbola, 280, 349. Triangle, area of, 186 ; angles of, 210 ; solution of, 181 ff., 244 ff. Trigonometric equations, 174. Trigonometric functions, definitions of, 147, 168 ; graphs of, 158, 159, 161, 167, 173, 193, 194; variation of, 157 ff. ; computation of, 148 ff., 152 ff. ; periods of, 157, 159, 162 ; inverse, 192 ff. ; formulas, 179, 180, 181, 201, 202, 204, 205, 207, 446; application of De Moivre's theorem, to expansion of, 446 ; logarithms of, 242, 538 ff. ; tables of, 538 ff. Trisection of an angle, 388. Two-valued function, 20, 265 ff. Variable, definition of, 12 ; independ- ent and dependent, 12 ; range of, 23. Vector, definition of, 5, 434 ; addition of — s, 435 ; components of, 436. Vectorial angle, 163. Versed sine, 168. Vertex of a conic, 109, 343, 349. Zero of a function. 455. Printed in the United States of America. n^HE following pages contain advertisements of books by the same author or on kindred subjects. TRIGONOMETRY BY ALFRED MONROE KENYON Professor of Mathematics in Purdue University And LOUIS INGOLD Assistant Professor of Mathematics in the University of Missouri Edited by Earle Raymond Hedrick Trigonometry, fie xible cloth, pocket size, long izmo {xi-\- ij2 pp.) with Complete Tables {xviii -\- 124 pp.) , $1.50 Trigonometry {xi-{- ij2 pp.) with Brief Tables {xviii -\- 12 pp.), $1.20 Macmillan Logarithmic and Trigonometric Tables, fiexlble cloth, pocket size, long i2mo (xviii + 124 pp.), $j6o FROM THE PREFACE The book contains a minimum of purely theoretical matter. Its entire organization is intended to give a clear view of the meaning and the imme- diate usefulness of Trigonometry. The proofs, however, are in a form that will not require essential revision in the courses that follow. . . . The number of exercises is very large, and the traditional monotony is" broken by illustrations from a variety of topics. Here, as well as in the text, the attempt is often made to lead the student to think for himself by giving suggestions rather than completed solutions or demonstrations. The text proper is short; what is there gained in space is used to make the tables very complete and usable. Attention is called particularly to the com- plete and handily arranged table of squares, square roots, cubes, etc.; by its use the Pythagorean theorem and the Cosine Law become practicable for actual computation. The use of the slide rule and of four-place tables is encouraged for problems that do not demand extreme accuracy. Only a few fundamental definitions and relations in Trigonometry need be memorized; these are here emphasized. The great body of principles and processes depends upon these fundamentals; these are presented in this book, as they should be retained, rather by emphasizing and dwelling upon that dependence. Otherwise, the subject can have no real educational value, nor indeed any permanent practical value. THE MACMILLAN COMPANY Fublishers 64-66 Fifth Avenue New York THE CALCULUS BY ELLERY WILLIAMS DAVIS Professor of Mathematics, the University of Nebraska Assisted by William Charles Brenke, Associate Professor ol Mathematics, the University of Nebraska Edited by Earle Raymond Hedrick Cloth, semi-Jlexible, xxi -\- 383 pp. + Tables {63), ismo, $2.10 Edition De Luxe, flexible leather binding, India paper, $2. so This book presents as many and as varied applications of the Calculus as it is possible to do without venturing into technical fields whose subject matter is itself unknown and incomprehensible to the student, and without abandoning an orderly presentation i)f fundamental principles. The same general tendency has led to the treatment of topics with a view toward bringing out their essential usefulness. Rigorous forms of demonstra- tion are not insisted upon, especially where the precisely rigorous proofs . would be beyond the present grasp of the student. Rather the stress is laid upon the student's certain comprehension of that which is done, and his con- viction that the results obtained are both reasonable and useful. At the same time, an effort has been- made to avoid those grosser errors and actual misstatements of fact which have often offended the teacher in texts otherwise attractive and teachable. Purely destructive criticism and abandonment of coherent arrangement are just as dangerous as ultra-conservatism. This book attempts to preserve the essential features of the Calculus, to give the student a thcrough training in mathematical reasoning, to create in him a sure mathematical imagination, and to meet fairly the reasonable demand for enlivening and enriching the subject through applications at the expense of purely formal work that con* tains no essential principle. THE MACMILLAN COMPANY Publishers 64-66 Fifth Avenue New York GEOMETRY BY WALTER BURTON FORD Junior Professor of Mathematics in the University of Michigan And CHARLES AMMERMAN The William McKinley High School, St. Louis Edited by Earle Raymond Hedrick, Professor of Mathematics in the University of Missouri Plane and Solid Geometry, cloth, i2mo, 31 q pp., $1.23 Plane Geometry, cloth, i2mo, 213 pp., $ .80 Solid Geometry, cloth, i2mo, 106 pp., $ .80 STRONG POINTS I. The authors and the editor are well qualified by training and experi- ence to prepare a textbook on Geometry. II. As treated in this book, geometry functions in the thought of the pupil. It means something because its practical applications are shown. III. The logical as well as the practical side of the subject is emphasized. IV. The arrangement of material is pedagogical. V. Basal theorems are printed in black-face type. VI. The book conforms to the recommendations of the National Com- mittee on the Teaching of Geometry. VII. Typography and binding are excellent. The latter is the reenforced tape binding that is characteristic of Macmillan textbooks. " Geometry is likely to remain primarily a cultural, rather than an informa- tion subject," say the authors in the preface. " But the intimate connection of geometry with human activities is evident upon every hand, and constitutes fully as much an integral part of the subject as does its older logical and scholastic aspect." This connection with human activities, this application of geometry to real human needs, is emphasized in a great variety of problems and constructions, so that theory and application are inseparably connected throughout the book. These illustrations and the many others contained in the book will be seen to cover a wider 7-ange than is usual, even in books that emphasize practical applications to a questionable extent. This results in a better appreciation of the significance of the subject on the part of the student, in that he gains a truer conception of the wide scope of its application. The logical as well as the practical side of the subject is emphasized. ^ Definitions, arrangement, and method of treatment are logical. The defi- nitions are particularly simple, clear, and accurate. The traditional manner of presentation in a logical system is preserved, with due regard for practical applications. Proofs, both foimal and informal, are strictly logical. THE MACMILLAN COMPANY Publishers 64 G6 Fifth Avenue New York Analytic Geometry and Principles of Algebra BY ALEXANDER ZIWET Professor of Mathematics, the University of Michigan AND LOUIS ALLEN HOPKINS Instructor in Mathematics, the University of Michigan Edited by EARLE RAYMOND HEDRICK Cloth, via + j6g pp., appendix, answers, index, i2mo, $i.7S This work combines with analytic geometry a number of topics traditionally treated in college algebra that depend upon or are closely associated with geometric sensation. Through this combination it becomes possible to show the student more directly the meaning and the usefulness of these subjects. The idea of coordinates is so simple that it might (and perhaps should) be explained at the very beginning of the study of algebra and geometry. Real analytic geometry, however, begins only when the equation in two variables is interpreted as defining a locus. This idea must be introduced very gradu- ally, as it is difficult for the beginner to grasp. The familiar loci, straight line and circle, are therefore treated at great length. In the chapters on the conic sections only the most essential properties of these curves are given in the text ; thus, poles and polars are discussed only in connection with the circle. The treatment of solid analytic geometry follows the more usual lines. But, in view of the application to mechanics, the idea of the vector is given some prominence; and the representation of a function of two variables by contour lines as well as by a surface in space is explained and illustrated by practical examples. The exercises have been selected with great care in order not only to fur- nish sufficient material for practice in algebraic work but also to stimulate independent thinking and to point out the applications of the theory to con- crete problems. The number of exercises is sufficient to allow the instructor to make a choice. To reduce the course presented in this book to about half its extent, the parts of the text in small type, the chapters on soUd analytic geometry, and the more difficult problems throughout may be omitted. THE MACMILLAN COMPANY Publishers 64-66 Fifth Avenue New York Elements of Analytic Geometry BY ALEXANDER ZIWET Professor of Mathematics, the University of Michigan AND LOUIS ALLEN HOPKINS Instructor in Mathematics, the University of Michigan Edited by EARLE RAYMOND HEDRICK As in most colleges the course in analytic geometry is preceded by a course in advanced algebra, it appeared desirable to publish separately those parts of the authors' "Analytic Geometry and Principles of Algebra" which deal with analytic geometry, omitting the sections on algebra. This is done in the present work. In plane analytic geometry, the idea of function is introduced as early as possible; and curves of the form>' =/(x), where /(;c) is a simple polynomial, are discussed even before the conic sections are treated systematically. This makes it possible to introduce the idea of the derivative ; but the sections dealing with the derivative may be omitted. In the chapters on the conic sections only the most essential properties of these curves are given in the text ; thus, poles and polars are discussed only in connection with the circle. The treatment of solid analytic geometry follows more the usual lines. But, in view of the application to mechanics, the idea of the vector is given some prominence ; and the representation of a function of two variables by contour lines as well as by a surface in space is explained and illustrated by practical examples. The exercises have been selected with great care in order not only to fur- nish sufficient material for practice in algebraic work, but also to stimulate independent thinking and to point out the applications of the theory to con- crete problems. The number of exercises is sufficient to allow the instructor to make a choice. THE MACMILLAN COMPANY Publishers 64-66 Fifth Avenue New York SLIDE-RULE 11 ■ i I 1 i i (i) (a) (s) T1 !■. Z?' ! ^ I t I II I I D' (^) Directions A reasonably accurate slide-rule may be made by the student, for temporary practice, as follows. Take three strips of heavy stiff cardboard 1".3 wide by &' long; these are shown in cross-section in (1), (2), (3) above. On (3) paste or glue the adjoining cut of the slide rule. Then cut strips (2) and (3) accurately along the lines marked. Paste or glue the pieces together as shown in (4) and (5). Then (5) forms the slide of the slide-rule, and it will fit in the groove in (4) if the work has been carefully done. Trim off the ends as shown in the large cut. < CQ O Q O --E = = n. = CO - ~ — -.CO 02 ■ = = -—03 z: -JZ = — = E E = = — - lO E E =: CO;^ ^--CD E CO -: = ii;CO .;j = 1""^ :e = : _ — _, ■E E- -E : = - ■rti — ;i — — " _ _ ■ '_ ^ 1: - = ~T^ :: - ■_— - " _ — — - -— -- to - - -. -CO — — - = E = - =1— (M-i |— <M ^ -i CO _ = ^» %--^ = = - 1 ~:z::z. = = -E l=- E E - = r; = .: ZZZ- — E - " — — -- — — < m o G 14 DAY USE RETURN TO DESK FROM WHICH BORROWED LOAN DEPT. This book is due on the last date stamped below, or on the date to which renewed. Renewed books are subject to immediate recall. 7^^^^ n RECCW. wylO'8 4 ^^l cTftCKS FEB ^f^P -D JUN^ '^4- 1PM MftR 01966 91 JUNiO'BFi^ O'^ CD INTER-LI BRAR CDSN MAR 2 9 1968 " MAY 241 9 84 LD 21A-40m-4,'63 (D6471sl0)476B General Library University of California Berkeley ^ 372326 V^' * iw W'.-^ #• *■ UNIVERSITY OF CAUFORNIA UBRARY a %■ .<,.